UNIVERSITY  OF  CALIFORNIA 
AT   LOS  ANGELES 


ELEMENTS  OF  GEOMETRY 


GEORGE  C 


^ARDS,  Ph.B. 


Associate  Professor  oB^yiwiiEMiSTics  in  the  Univbrsity  of 


'  .'  1  t 


MACMILLAN    AND    CO. 

AND    LONDON 

1895 


All  rights  reserved 


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lilllt) 


COPTBIOHT,   1895, 

By  GEOBGE  C.  EDWARDS. 


) 


Nortoooti  H^xtvi 

J.  S.  Cnthing  &  Co.  —  Berwick  &  Smith 
Norwood  Masa.  U.S.A. 


w 


Engineering*  p>.    A 

Msthematical  ^^^ 

Sciences    4  5     '3 

Library        ' 


PREFACE. 


The  large  number  of  Geometries  recently  published 
is  an  indication  of  the  fact  that,  so  far  as  Geometry  is 
concerned,  there  is  an  unsatisfied  want  in  High  Schools, 
Academies,  and  Colleges.  It  is  in  the  hope  of  supplying 
this  need  that  the  book  here  presented  has  been  written. 

Effort  has  been  made  so  to  frame  the  definitions  that 
they  will  not  have  to  be  changed  when  the  student 
comes  to  the  higher  reaches  of  the  subject,  or  w"hen 
he  advances  to  other  branches  of  mathematics.  Effort 
has  been  made  to  introduce  new  material  when  needed, 
and  not  before ;  to  appeal  to  the  understanding ;  to  dis- 
courage mere  memorizing;  and  to  avoid  entering  into 
confusing  details. 

Corollaries  and  scholia  have  been  in  large  part  re- 
placed by  exercises,  every  one  of  which  the  student 
must  work  out  as  he  comes  to  them. 

At  the  end  of  the  Plane  Geometry  and  at  the  end  of 
the  Solid  Geometry  there  will  be  found  a  sufficiently 
large  number  of  exercises  to  give  a  review  of  the  work 
preceding  them,  and  thoroughly  to  establish  method  of 


VI  PREFACE. 

attack  in  the  mind  of  the  student.  The  exercises  which 
have  been  taken  from  other  authors  are  limited  to  such 
as  appear  in  at  least  two  of  them. 

Simplicity  in  wording  and  in  demonstration  have  been 
sought;  numerous  notes  are  added;  leading  thoughts, 
definitions,  and  methods  are  repeated  so  as  to  enforce 
them ;  the  student  is  led,  not  driven ;  and  he  is  encour- 
aged to  investigate  and  determine  for  himself. 

The  division  of  the  work  into  fourteen  chapters  makes 
a  natural  arrangement  of  parts,  although  the  chapters 
are  quite  unequal  in  length.  The  chapters  have  been 
divided  into  articles,  and  these  have  been  numbered  for 
convenience  in  cross-reference.  But  the  student  should 
refer  to  principles  or  facts  by  stating  them  in  full. 

The  writer  thinks  that  the  proper  place  for  the  theory 
of  proportion  is  in  the  Algebra,  and  for  that  reason  has 
omitted  it  here. 

The  leading  features  of  the  book  are,  the  development 
of,  and  the  insistence  on,  method  of  attack  in  the  solution 
of  problems. 

Logic,  which  is  the  science  of  orderly  thinking, 
requires  for  its  proper  study,  a  development  of  mind 
beyond  that  which  is  necessary  for  the  student  of  the 
ordinary  Geometry.  Geometry,  however,  does  admirably 
illustrate  the  principles  of  Logic ;  and  in  it  these  princi- 
ples are  rigidly  enforced  at  every  step  without  the  dis- 
tractions that  accompany   investigations   in   any  other 


PEEFACE.  Vll 

field  of  research.  From  the  educational  point  of  view 
Geometry  is  the  most  valuable  study  of  the  High  School 
period. 

Instructors  are  advised  to  proceed  slowly  with  the 
earlier  chapters,  and  more  rapidly  with  the  later  ones; 
not  abating  at  all  in  thoroughness. 

Instructors  and  students  are  advised  to  read  carefully 
the  paragraphs  on  Demonstrative  Geometry,  pages  112- 
116  of  the  "Eeport  of  the  Committee  on  Secondary 
School  Studies,  appointed  at  the  meeting  of  the  National 
Educational  Association,  July  9,  1892 " ;  and  published 
by  the  United  States  Bureau  of  Education  in  1893. 

The  writer  will  deem  it  a  favor  to  have  his  attention 
called  to  errors.  They  will  surely  exist  in  the  first 
writing,  but  the  author  hopes  that  the  main  features 
of  the  book  will  not  be  seriously  injured  thereby. 

GEORGE  C.  EDWARDS. 
Berkeley,  March  23,  1895. 


CONTENTS. 


PAGE 

Intkoduction     .        . x^ 

CHAPTEE  I. 

Definitions 1 

Direction  axiom 2 

Rotation  axiom r  4 

An  angle 5 

Figures 6 

Translation  axiom 6 

Equality  axioms 12 

Perpendiculars 14 

CHAPTER  II. 

A  triangle 17 

Parallels 21 

Perpendicular  bisector 24 

A  circle 26 

Congruent  triangles 26 

Construction  of  triangles 28 

Inequality  axioms 32 

CHAPTER  III. 

Triangles 36 

Quadrilaterals  and  quadrangles 42 

ix 


X  CONTENTS. 

PAGK 

Polygons 46 

Analysis 48 


CHAPTER  IV. 

A  circle 61 

Angles  and  arcs 64 

Triangles  and  circles 64 

CHAPTER   V. 

Measurement  of  distance 69 

Area 71 

Proportional  division 76 

Medians 83 


CHAPTER  VI. 

Squares  on  segments 87 

Squares  on  the  sides  of  triangles 89 

Areas  of  similar  triangles 95 

Areas  of  similar  polygons ~96 

Metrical  relations  of  parte  of  a  triangle 99 


CHAPTER  Vn. 

Chords  and  tangente 103 

CHAPTER  VIIL 

Inscribed  and  circumscribed  polygons 117 

Variable  and  limit 121 

Limit  axioms 124 

Area  of  a  circle 126 

Pboblkus 137 


CONTENTS.  xi 


CHAPTER   IX. 

PAGK 

Intersections  of  planes 175 

Perpendiculars  to  planes 177 

Parallels  to  planes 184 


CHAPTER   X. 

A  sphere 188 

Plane  sections  of  a  sphere 188 

Spherical  arcs 189 

Spherical  triangles 192 

CHAPTER   XI. 

Triedrals 205 

CHAPTER   XII. 

Prisms 210 

Pyramids 210 

Cylinders 213 

Cones 219 

Surface  of  a  sphere 227 

Zone 229 

Lune 230 

Area  of  a  spherical  triangle 231 

Areas  of  similar  surfaces 232 

CHAPTER   XIII. 

Volume  of  a  prism 236 

Volume  of  a  pyramid 241 

Prismoidal  formula 245 

Truncated  prism 246 

Volume  of  a  sphere 249 

Volumes  of  similar  figures 251 


XU  CONTENTS. 


CHAPTER  XIV. 

PAGE 

The  parabola 264 

Line  relations 264 

Area  of  a  parabola 262 

Particular  cases 264 

The  ellipse 265 

Line  relations 266 

Area  of  an  ellipse 270 

Particular  cases 272 

The  hyperbola 272 

Line  relations 274 

Particular  cases .  281 

Problems 283 


SYMBOLS. 


Z,  Angle. 

A,  Angles. 

A,  Plane  triangle. 

A,  Plane  triangles. 

= ,  Equals,  or  is  equal  to. 

II,  Parallel. 

<,  Less  than. 

>,  Greater  than. 

O,  Parallelogiam. 

m,  Parallelograms. 

±,  Perpendicular. 

O,  Circle,  or  circumference  (depending  on  the  context). 

sp.  A,  Spherical  triangle. 

PQ,  The  straight  line  segment  FQ. 

PQ,  The  arc  PQ. 

Q.  E.  D.,  Quod  Erat  Demonstrandum  (which  was  to  be  proved). 

Q.  E.  F.,  Quod  Erat  Faciendum  (which  was  to  be  done). 

.'.,  Hence. 


INTRODUCTION. 


Geometky  had  its  origin,  as  the  name  indicates,  in  the 
need  of  a  method  for  the  accurate  description  of  limited 
portions  of  the  Earth's  surface. 

At  the  present  time  the  development  of  the  science  is 
such  as  to  include  a  vast  amount  of  material  which  is 
beyond  the  scope  of  a  book  devoted  to  the  consideration 
of  elementary  forms  in  such  space  as  that  in  which  we 
live  and  exercise  our  senses. 

Like  any  other  science,  Geometry  is  built  upon  defi- 
nitions and  axioms.  Definitions  describe,  in  as  simple  a 
manner  as  possible,  the  objects  with  which  we  have  to 
deal.  Axioms  are  self-evident  truths  that  relate  to  the 
objects  described,  and  the  operations  to  be  performed. 
An  axiom  is  a  truth  self-evident  to  one  who  imderstands 
the  terms  in  which  it  is  stated. 

Upon  the  definitions  and  axioms  of  Geometry  is  built 
up  our  knowledge  of  fundamental  relations.  Upon  these 
more  complex  relations  are  based.  Upon  these  new 
relations,  still  more  advanced  relations  are  built;  and 
thus  the  science  is  developed.     As  is  true  of  every  other 


XVI  INTRODUCTION. 

science,  the  extent  to  which  Geometry  may  be  pursued 
is  without  limit. 

While  tKe  beginnings  are  based  upon  definitions  and 
axioms,  it  does  not  follow  that  all  of  the  definitions  and 
axioms  are  to  be  brought  forward  at  the  opening  of  the 
subject.  As  the  information  of  the  student  grows,  new 
definitions  and  new  axioms  are  appropriate. 

Advance  in  Geometry  is  made  through  an  orderly 
arrangement  of  theorems  and  problems. 

A  theorem  is  a  general  statement  of  relations.  These 
relations  are  established  by  a  course  of  reasoning. 

A  problem  is  a  demand  that  a  construction  be  made; 
that  certain  relations  be  established;  or  that  relations 
between  certain  things  be  established. 


ELEMENTS  OF  GEOMETRY. 


o>«<c 


CHAPTER   I. 

1.  Definitions.  An  enclosed  portion  of  space,  no  matter 
what  its  form  or  material,  or  if  it  exist  in  the  imagina- 
tion only,  is  called  a  volume. 

That  which  separates  a  volume  from  excluded  space 
is  called  surface,  and  does  not  partake  of  the  character 
of  the  material,  if  there  be  any  used.     The  surface  that 


Fig.  1. 

separates  the  hull  of  a  ship  from  the  water  is  neither 
water  nor  the  material  of  which  the  ship  may  be  con; 
structed. 

2.  Definitions.  A  position  in  space  that  is  without 
magnitude  is  called  a  point. 

If  a  point  move  it  will  generate  a  line ;  it  may  move 
at  a  snail's  pace,  or  it  may  move  with  the  rapidity  of 
thought. 


2  ELEMENTS   OF   GEOMETRY. 

The  position  of  one  point  with  respect  to  another 
determines  direction.  If  we  represent  the  position  of 
one  point  by  the  letter  (^4),  placed  near  it,  and  the 
position  of  another  point  by  the  letter  (B),  placed  near 
it,  the  direction  (AB)  is  established.  The  direction  of 
(B)  from  (A)  and  the  direction  of  (A)  from  (B)  are 
opposites. 

Relations  expressed  by  the  words  same  and  opposite, 
whether  they  be  of  interpretation  or  of  operation,  are 
expressed  symbolically  by  the  vertical  cross  (+),  and  by 
the  horizontal  dash  (— )•  If  (+4)  represents  a  number 
of  miles  in  the  direction  of  (B)  from  (A),  (—4)  will 
represent  the  same  distance  in  the  direction  of  {A)  from 
(B).  If  (+6)  represents  six  years  to  come,  (—4)  will 
represent  four  years  that  have  passed. 

Anything  which  is  so  large  that  it  is  beyond  our  com- 
prehension, is  said  to  be  infinitely  large.  Space  is  infir 
nitdy  large;  the  number  of  points  in  space  is  infinite; 
time  is  infinite. 

Direction  Axiom.  From  any  point  in  space  there  tcill 
be  an  infinite  number  of  directions;  and  each  direction  will 
have  its  opposite  direction. 

There  will  also  be  an  infinite  number  of  points  in 
any  assumed  direction  from  a  given 
point,  which  points  will  each  be  at 
different  distances  from  the  given 
point.  In  each  opposite  direction 
there  will  also  be  an  infinite  number 
of  points.     The  rays  of  light  from  F'«-  2. 

an  arc-light  or  from  a  fixed  star  are  fairly  good  illus- 
trations. 


DEFINITIONS.  8 

3.  If  a  point  move  from  an  assumed  position,  it  may 
start  ill  any  one  of  an  iniinite  number  of  directions.  If 
it  continue  in  the  direction  in  which  it  starts,  so  as  to 
pass  through  the  position  of  every  point  lying  in  the 
same  direction  (§  2),  it  will  generate  what  is  called  a 
straight  line. 

Since  a  straight  line  is  determined  by  direction,  and 
two  points  determine  direction,  two  points  determine  the 
position  of  a  straight  line. 

If  separate  straight  lines  have  any  point  in  common 
there  can  be  but  one,  for  if  they  have  two  coincident 
points  the  two  lines  will  form  one  line. 

The  straight  line  is  infinite  in  extent. 

The  portion  of  a  straight  line  between  any  two  points 
of  the  line  is  called  a  segment. 


B 

Fig.  3. 


Unless  otherwise  specified,  the  indefinite  straight  line 
is  to  be  understood  when  two  points  are  named  which 
locate  the  line,  as :  the  line  AB  (the  segment  is  repre- 
sented by  AB). 

An  infinite  number  of  straight  lines  may  pass  through 
a  point. 

If  a  point  move  with  ever-changing  direction,  it  will 
generate  a  curved  line  ;  the  law  of  change  determining 
the  character  of  the  curve. 

As  any  fixed  point  has  a  fixed  direction  from  another 
fixed  point,  only  one  straight  line  can  join  two  points. 


4  ELEMENTS   OF   GEOMETRY. 

4.  Our  idea  of  distance  involves  the  ideas  of  motion 
and  of  time.  Time  is,  and  advances  with  uniformity. 
Motion  may  or  may  not  be  uniform.  If  motion  be  uni- 
form, the  greater  the  time  the  greater  the  distance.  In 
equal  times  the  more  rapid  of  two  motions  will  cover  the 
greater  distance. 

If  a  point  move  from  position  A  to  position  B  without 
change  of  direction,  it  will  reach  position  B  after  having 
moved  over  a  less  distance,  than  if  on  the  way  it  had 
made  any  changes  of  direction. 

We  therefore  say  that  the  shortest  distance  between 
two  points  is  that  portion  of  the  straight  line  determined 
by  them  (§  3)  which  lies  between  them. 

Although  the  straight  lines  with  which  we  deal  are  in 
general  infinite  in  length,  our  representations  of  them 
are  limited  by  the  extent  of  the  surface  upon  which  our 
representations  are  made. 

Henceforth  when  the  word  line  is  used  without  any 
qualification  a  straight  line  is  understood. 

5.  Rotation  Axiom.  If  two  straight  lines  intersect, 
either  or  both  may  be  moved  without  changing  the  point  of 
intersection.  They  may  be  brought  to  coincide;  in  ivJiich 
case  all  the  points  of  the  two  lines  are  in  common.  The 
lines  are  then  said  to  be  congruent. 


PLANE  GEOMETRY. 


6.  A  plane  is  a  surface  such  that  if  any  two  of  its 
points  are  joined  by  a  straight  line  it  will  contain  the 
entire  line. 

An  infinite  number  of  straight  lines  may  lie  in  a  plane. 

In  drawing,  a  dot  upon  a  plane  is  used  as  the  repre- 
sentative of  a  point,  and  a  mark  by  chalk,  pencil,  or  pen 
is  used  to  represent  a  line. 

AN  ANGLE. 

7.  Let  the  surface  of  this  page  represent  a  plane,  and 
AB  a  line  of  the  plane. 


Fig.  4. 


If  the  direction  of  B  from  A  be  positive  (+),  the  direc- 
tion of  C  from  A  will  be  negative  (— ). 

If  the  straight  line  rotate  about  the  point  ^  as  a  pivot, 
and  remain  in  the  plane  of  the  page,  it  will  make  a 
change  in  direction.  If  the  motion  be  considered  as 
having  stopped  when  the  line  has  arrived  at  the  position 
AD,  the  change  in  direction  is  called  an  angle ;  and  A  is 
called  its  vertex. 


5 


6  ELEMENTS   OF  GEOMETRY. 

The  angle  may  be  described  as:  the  angle  BjLD 
(Z  BAD),  or  by  a  single  symbol  as  6. 

The  distances  AB  or  AD  have  nothing  whatever  to  do 
with  the  magnitude  of  the  angle ;  it  is  purely  a  matter 
of  direction. 

FIGURES. 

8.  Volumes,  surfaces,  lines,  angles,  and  points,  or  any 
combination  of  them  that  we  may  make,  constitute  geo- 
metric figures. 

The  relations  of  parts  of  the  same  figure  to  each  other, 
and  the  relations  of  different  figures  to  each  other,  con- 
stitute the  subject  matter  of  geometry. 

The  Elementary  Geometry  is  ordinarily  separated  into 
two  parts,  viz. : 

(a)  That  which  relates  to  figures  in  a  plane,  and  is 
called  Plane  Geometry. 

(6)  That  which  relates  to  figures  that  do  not  lie  in  one 
plane  only,  and  is  called  Sohd  Geometry. 

From  now  on  until  we  arrive  at  §  105,  we  shall  be  con- 
cerned with  Plane  Geometry. 

Translation  Axiom.  A  geometric  figure  may  he 
moved  at  pleasure  without  changing  the  relations  existing 
between  the  parts  which  compose  it. 

MORE  ABOUT  PLANES. 

9.  A  plane  may  be  made  to  pass  through  a  given 
point :  for  a  plane  may  be  moved  so  as  to  cause  any  one 
of  its  points  to  coincide  with  the  given  point. 


MORE   ABOUT   PLANES.  7 

Any  infinite  number  of  planes  may  be  made  to  pass 
through  a  given  point:  because  the  manner  of  moving 
any  plane  so  as  to  cause  it  to  contain  a  given  point  has 
not  been  limited. 

If  a  plane  pass  through  a  point,  any  line  which  con- 
tains the  given  point,  and  lies  in  the  plane,  may,  by 
motion  of  the  plane,  be  brought  to  pass  through  a  second 
point,  located  anywhere. 

Thus  Ave  see  that  a  plane  may  be  passed  through  any 
two  points.  It  will  contain  the  straight  line  joining 
those  points.  Every  possible  position  of  a  plane  which 
contains  two  given  points,  may  be  reached  by  causing  a 
selected  plane  which  contains  them  to  rotate  about  the 
line  joining  them  as  an  axis,  until  it  returns  to  the 
position  it  had  at  starting.  Thus  we  see  that  an  infinite 
number  of  planes  may  be  passed  through  two  points. 
These  planes  will  each  contain  the  line  determined  by 
the  two  points. 

A  plane  passed  through  a  line  and  making  a  complete 
rotation  on  the  line  as  an  axis,  will  encounter  every  point 
in  space. 

10.  If  a  line  in  each  of  two  planes  be  made  to  coincide, 
and  then  one  of  the  planes  be  rotated  about  this  line 
until  at  least  07ie  point,  not  in  the  common  line,  shall  be 
common  to  the  two  planes,  any  line  which  may  be  drawn 
through  this  point  and  any  point  of  the  common  line 
will  lie  in  both  planes.  Lines  may  therefore  be  drawn 
through  this  point  so  as  to  reach  every  point  of  both 
planes ;  in  which  case  the  two  planes  would  have  every 
point  in  common.     Hence  the  statement  or 


8  ELEMENTS   OF   GEOMETRY. 

Theorem.     Two  planes  may  he  made  to  coincide. 

Bemark.  —  By  virtue  of  this  possibility  all  planes  are  said  to  be 
congruent. 

11.  If  the  line  AB  remaiu  in  the  plane  of  the  page, 
and  be  revolved  about  ^  as  a  pivot,  it  will  generate  an 
angle  by  its  change  of  direction  (§  7).  When  the  posi- 
tive direction  AB  has  changed  to  that  of  AC,  the  angle 
BAC  (Z  BAC)  will  have  been  generated.  When  it  shall 
have  been  still  further  changed  to  the  direction  AD, 
Z  BAD  will  be  the  resvdt.  The  rotation  may  still  con- 
tinue, and  the  positive  direction  may  become  in  succes- 


Fio.  6. 


sion,  AE,  AF,  AG,  AH,  and  may  finally  coincide  with 
its  first  position,  AB. 

If  the  direction  AB  be  positive,  the  direction  AF  is 
negative.  "When  the  positive  direction  has  been  changed 
to  AC,  the  negative  has  been  changed  to  AI.  When  the 
positive  has  been  changed  to  AD,  the  negative  has  been 


AFGLES.  y 

changed  to  AJ.  When  the  positive  has  arrived  at  AF, 
the  negative  has  arrived  at  AB.  And  when  the  positive 
has  again  come  to  the  position  AB,  the  negative  will 
have  returned  to  its  corresponding  position,  AF. 

Both  the  positive  and  the  negative  directions  have 
generated  angles,  and  each  has  made  a  complete  rotation. 
In  this  rotation  each  has  passed  through  every  point  in 
the  plane.  Since  all  positions  are  thus  merely  dupli- 
cated, the  angle  generated  by  the  positive  direction  is, 
in  general,  the  only  one  considered. 

For  convenience  of  description  and  measurement  the 
complete  rotation  is  separated  into  parts ;  generally  into 
360  equal  parts,  each  being  called  a  degree;  sometimes 
into  400  equal  parts,  each  being  called  a  grade, 

Degrees  are  again  separated  into  60  equal  parts,  each 
called  a  minute ;  each  minute  is  separated  into  60  equal 
parts,  called  seconds ;  and  each  second  is  then  separated 
decimally,  if  a  further  division  is  necessary. 

Grades  are  separated  decimally. 

Each  system  of  division  and  subdivision  has  its  advan- 
tages. The  sexagesimal  is  the  system  in  common  use, 
and  has  been  since  a  period  of  time  long  antedating  the 
Christian  era.  Nearly  all  of  the  literature  and  instru- 
ments of  observation  involving  the  consideration  of 
angles  employ  this  system. 

But  the  decimal  system  accords  with  our  method  of 
enumeration,  and  will  at  some  time  supplant  the  other. 
May  the  time  soon  come. 

If  the  portion  of  the  plane  below  the  line  FB  be 
revolved  on  that  line  as  an  axis,  it  may  be  brought  to 
coincide  with  that  which  is  above  the  same  line  (§  10) ; 
and  in  that  position,  if  the  line  AB  be  rotated  about  A 


10  ELEMENTS   OP   GEOMETRY. 

as  a  pivot  until  it  reach  the  direction  AF,  it  will  have 
generated  the  same  angle  in  the  two  coincident  ligures. 
Hence  the  two  angles  thus  generated  must  be  equal. 
Each  is  one-half  of  the  angle  generated  by  a  complete 
rotation ;  and  may  be  expressed  as  180  degrees  (180°),  or 
200  grades  (200«). 

The  dotted  line  in  the  accompanying  figure  is  used  to 
indicate  the  rotation  by 

marking  the  path  of  the  ^-^°->. 

point  B  during  the  gen- 
eration of  the  angle  of 
360°,  which  is  separated 
into  two  equal  parts  by 
the  line  FB. 

If  the  line  AB  change 
its  direction  90°,  it  will 
occupy  the  position  in- 
dicated by  AD,  and  the  angles  BAD  and  DAF  will  be 
equal. 

If  the  part  of  the  figure  to  the  right  of  the  line  AD 
be  revolved  on  AD  as  an  axis  until  the  revolved  portion 
coincides  with  the  part  on 


/ 

/ 

/ 

\ 

\ 
\ 

1 

< 

A                ,B 

\ 

\ 

J 

180° 

Fig.  6. 

D 


F 


B 


the  left  of  AD,  the  Z 
DAB  will  coincide  with 
the  Z  DAF;  ^5  will  fall 
in  the  direction  AF;  the 
Z  BAJ  will  coincide  with 
ZFAJ;  and  so  must  be 
equal  to  it. 

Hence  (.-.)  the  180°  of 
change  of  direction  from  ^"'-  '^• 

AF  to  AB  will  be  bisected  by  AJ.     Thus  we  see  that  if 


(30) 


ANGLES. 


11 


two  lines  intersect  so  as  to  form  an  angle  of  90°,  there 
will  be  formed /owr  angles  of  90°  each. 

An  angle  of  90°  is  sometimes  called  a  quadrant,  but 
more  frequently  a  right  angle. 

Lines  at  right  angles  with  each  other  are  said  to  be 
perpendicular  (_L) ;  and  either  line  with  respect  to  the 
other  is  called  a  perpendicular. 

When  the  sum  of  two  angles  is  90°,  each  is  said  to  be 
the  complement  of  the  other. 

When  the  sum  of  two  angles  is  180°,  each  is  said 
to  be  the  supplement  of  the  other. 

An  angle  that  is  less  than  90°  is  said  to  be  acute. 

An  angle  greater  than  90°,  and  less  than  180°,  is  said  to 
be  obtuse. 

Acute  and  obtuse  angles  are  frequently  spoken  of  as 
oblique. 

1, 


Fig.  8. 

Angles  which  occupy  toward  each  other  the  position 
that  A  AOB  and  EOF  do,  are  said  to  be  vertical.  A  AOF 
and  BOE  are  vertical. 

Note. — Although  straight  lines  and  planes  both  extend  to  in- 
finity, it  is  only  the  relations  between  parts  at  a  finite  distance  that 
concern  us  in  the  present  work. 


12  ELEMENTS   OP   GEOMETRY. 

12.   The  sign  of  equality  is  (=). 

In  considering  the  aggregation  of  quantities,  addition 
is  indicated  by  (+),  and  subtraction  by  (— ),  (§  2). 

Equality  Axioms,     (a)   If  two  things  are  eqvM  to  a 
third  thing,  they  are  equal  to  each  other. 

(b)  The  whole  is  greater  than  any  of  its  parts,  and  equals 
the  sum  of  all  its  parts. 

(c)  If  the  same  operation  he  performed  upon  equals,  the 
results  ivill  be  equal. 

Theorem.     If  two  lines  intersect,  the  vertical  angles  are 
equal. 


Fig.  9. 

Proof  ZBAC+Z  CAF  =  180°,  §  11 . 

Z  CAF  +  Z  FAG  =  180°.  §  11. 

By  Equality  Axiom  (a), 

Z  BAC  +  Z  CAF  =  Z  CAF  +  Z  FAG. 
Then  subtracting  Z  CAF  from  each  member  of  the 
equation,  acting  imder  the  authority  given  by  Equality 
Axiom  (c),  we  have :  Z  BAC  =  Z  FAG.  q.  e.  d. 

Again,  Z  BAC  +  Z  CAF  =  180°, 

Z  BAC  +  Z  BAG  =  180°. 
By  Equality  Axiom  (a), 

Z  BAC  +  Z  CAF  =  Z  BAC  +  ZBAG. 
Then  subtracting  Z  BAC  from  each  member  of  the 
equation,  we  have :  Z  CAF  =  Z  BAG.  q.  e.  d. 


ANGLES. 


13 


ci 

I 


Exercise.  — Establish  the  fact  that  Z  BAC  =  Z  FAG,  by  rotat- 
ing one  of  them  until  it  coincides  with  the  other. 

Proof.  —  Under  the  authority  given  by  the  axiom  in  §  8,  the 
ZBAG  may  (without  changing  the  relations  between  its  parts)  be 
rotated  abovit  J.  as  a  pivot.  When  the  line  AB  shall  have  rotated 
180°,  the  line  AC  will  also  have  rotated  180°.  AB  will  then  coin- 
cide with  AF,  and  AC  with  AG. 

Therefore  all  the  parts   coincide  and  the  Z  FAG  equals  the 

BAG.  Q.E.D. 

Note.  —  A  mechanical  device  for  determining  approximately 
the  value  of  an  angle,  and  for  the  purpose  of  moving  a  given 
angle  to  any  desired  position,  is  called  a  protractor.  A  con- 
venient form  for  paper  or  blackboard  work  may  be  made  from 
pasteboard. 
I    Some  of  the  forms  used  are : 


Fig.  10. 


14 


ELEMENTS   OF   GEOMETRY. 


B 


13.   Theorem.     At  a  given  point  in  a  line  one,  and 
only  one,  perpendicular  can  he 
erected. 

(a)  We  may  at  any  point 
(A)  of  a  line  have  at  least 
one  perpendicular :  because 
any  point  of  a  line  might 
serve  as  a  pivot  about  which 
rotation  could  take  place,  and  Fig.  ii. 

when  the  rotation  has  extended  to  one-fourth  of  a  com- 
plete rotation  it  Avould  be  a  perpendicular  (§  11). 

(b)  If  two  perpendiculars,  as  AG  and  AD,  could  be 
erected  at  the  same  point,  we  should  have  two  unequal 
angles,  each  equalling  90°,  which  by  our  axiom  cannot  be. 

Hence  the  Theorem  is  established. 

Note. — The  T  square  is  a  me- 
chanical device,  used  in  drawing, 
for  the  purpose  of  erecting  perpen- 
diculars at  any  points  of  a  line ; 
and,  as  we  shall  see  in  the  next 
article,  for  letting  fall  perpen- 
diculars to  a  line  from  points  not 
in  the  line. 


Fio.  12. 


14.  Theorem.  From  a  point  not  on  a  given  straight 
line  one,  and  only  one,  perpendicular  may  be  drawn  to  the 
line. 

(a)  Since  at  any  point  of  a  line  a  perpendicular  can  be 
erected  to  the  line  (§  13),  we  may  conceive  of  a  perpen- 
dicular as  moving  along  the  line  so  as  to  be  always 
perpendicular  to  it. 

The  moving  perpendicular  may  be  caused  to  pass 
through  any  point  in  the  plane,  as  P;  when  it  does  so 


PERPENDICTJLARS. 


15 


B 


Fig.  13. 


pass,  there  will  be  at  least  one  perpendicular  to  the  line 
AB  through  the  point  P. 

(b)  Let  BC  represent  the  given  line,  and  FA,  a  per- 
pendicular through  A. 

If  any  other  straight  line 
could  be  drawn  through  A 
that  should  be  perpendic- 
ular to  BC,  let  AE  repre- 
sent it. 

On   the   ±  FA,  lay   ofE 
FB  =  AF,  and  draw  DE. 
AE  and  DE  will  intersect  and  have  but  one  point  in 
common  by  the  direction  axiom. 

If  AE  is  a  perpendicular,  DE  will  also  be  one :  because 
the  figure  below  the  line  BC  may  be  revolved  on  BG 
as  an  axis,  and  be  made  to 
coincide  with  that  portion 
of  the  figure  above  the  line ; 
the  angles  remaining  the 
same  during  the  revolution. 

Then  if  AEB  is  a  right 
angle,  DEB  will  be  one. 
But  by  §  11,  if  AEB  is  a 
right  angle,  KEB  will  also 
be  a  right  angle. 

And  KEB  would  equal 
DEB,  which  is  in  conflict  with  axiom  (6)  of  §  12.    ^  "^ 

Hence  the  theorem  is  established. 

Note.  —  The  supposition  that  there  could  be  a  perpendicular 
through  A,  other  than  AF,  to  the  line  BC  leads  to  conclusions 
which  are  in  conflict  with  our  axioms.  Hence  the  supposition  has 
been  shown  to  be  an  erroneous  one. 


Fig.  14. 


16  ELEMENTS  OF   GEOMETRY. 

The  student  will  carefully  note  the  method  employed  in  §§  13 
and  14,  It  is  shown  that  a  perpendicular  through  A  may  exist, 
and  that  another  line  through  A,  which  shall  be  perpendicular  to 
BC,  cannot  be  drawn. 

This  method  is  called  '^^Beductio  ad  absurdum." 

Exercise.  —  Use  the  above  figure  to  establish  the  fact  that  the 
least  distance  from  any  point  to  a  straight  line  will  be  the  segment 
AF  of  the  perpendicular. 

This  last  distance  named  is  the  one  always  to  be  imderstood  as 
the  distance  of  a  point  from  a  line,  or  as  the  distance  of  a  line  from 
a  point. 


CHAPTER   II. 
A  Triangle. 

15.  Definitions.  The  figure  formed  by  three  straight 
lines  which  intersect  so  as  to  enclose  a  portion  of  a  plane 
is  called  a  triangle  (A). 

The  points  of  intersection 
of  the  lines  forming  the  tri- 
angle are  called  the  vertices. 

At  each  vertex  four  an- 
gles are  formed.  The  angles 
within  the  enclosure  are  called 

.    ,     .  Fig.  15. 

intenor. 

At  each  vertex  there  is  an  angle  vertical  to  the  interior 
one,  and  equal  to  it.  There  is  also  at  each  vertex  a 
pair  of  vertical  angles,  each  supplementary  to  the  interior 
angle. 

The  segments  AB,  BC,  and  CA  are  called  sides  of  the 
angle.     Taken  together  they  are  called  the  perimeter. 

Note.  — The  figure  called  a  triangle  might  with  equal  propriety 
be  called  a  trilateral. 

16.  If  a  point  from  a  position  {S)  on  the  side  AB, 
move  in  the  directions  indicated  by  the  arrow-heads,  and 
return  to  S  after  having  moved  along  the  perimeter  of 
the  triangle,  it  will  have  changed  direction  at  B,  at  C, 
and  at  A ;  and  not  at  any  other  point.  At  B  it  would 
change  direction  to  the  left,  an  amount  indicated  by  Z  1 ; 

c  17 


Fig.  16. 


18  ELEMENTS   OF   GEOMETRY. 

at  C,  it  would  change  direction  to  the  left,  an  amount 
indicated  by  Z  2,  and  at  A  it  would  change  direction  to 
the  left,  an  amount  indi- 
cated by  Z  3. 

This  appears  to  be  a  total 
change  of  direction  equal 
to  a  complete  rotation. 

If  it  be  such,  we  should 
be  able  to  rotate  some  seg- 
ment of  a  straight  line  about 
one  of  its  extremities  and  have  its  partial  rotations  coin- 
cide with  the  changes  of  direction,  and  its  complete 
rotation  be  the  sum  of  the  three  changes  of  direction. 

Let  {ST)  be  any  segment  of  one  of  the  lines  forming 
the  triangle.  Move  it  in  the  line,  of  which  it  is  a  part, 
until  the  point  8  coincides  with  B.  Kotate  the  segment 
until  it  shall  coincide  in  direction  with  BC.  Then  move 
it  in  the  line  BG  until  the  point  S  comes  to  C.  Then 
rotate  until  the  direction  CA  is  taken  up.  Along  this 
line  translate  the  segment  until  the  point  S  shall  coincide 
with  A;  then  rotate  until  the  direction  AB  is  reached. 
Along  this  line  move  the  assumed  segment  until  it  shall 
have  come  to  its  initial  position. 

The  motion  of  the  segment  has  been  of  two  kinds: 
translation  and  rotation.  The  sum  of  the  translations 
has  been  the  perimeter  of  the  triangle,  and  the  sum  of  the 
rotations  has  been  a  complete  rotation  (360°  or  4  rt.  A). 
Hence  the 

Theorem.  Tlie  sum  of  the  changes  of  direction  {called 
exterior  angles)  equals  4  right  angles. 

Notes.  —  1.  If  from  the  point  S  we  should  proceed  in  the  oppo- 
site direction,  the  changes  of  direction  at  the  points  A,  C,  and  B 


A  TRIANGLE.  19 

would  be  right  handed,  and  would  be  the  verticals  of  angles  3,  2, 
and  1 ;  and  the  sum  would  again  be  4  right  angles. 

2.  The  student  will  observe  that  from  a  single  triangle  we  have 
determined  a  relation  which  we  say  is  true  of  all  triangles.  The 
reason  for  this  is  :  we  have  drawn  our  conclusions  without  putting 
any  limitations  upon  the  triangle.  Any  triangle  might  have  the 
same  method  applied  to  it  with  exactly  the  same  result.  We  are 
then  not  drawing  general  conclusions  from  special  cases. 

"     Exercises.  —  1.    Show  that  the  sum  of  the  interior  angles  of  a 

triangle  =  180°. 

L  2.    Show  that  any  angle  of  a  triangle  will  lie  between  0°  and 

180°. 
C-  3.    Show  that  but  one  angle  of  a  triangle  may  be  90°  or  greater. 

4.  Show  that  the  sum  of  two  sides  of  a  triangle  is  greater  than 
the  third. 

5.  Given  two  angles  of  a  triangle  to  find  the  third. 

17.  Theorem.  If  two  straigJit  lines  make  equal  angles 
with  a  third  straight  line  intersecting  them,,  they  will  make 
equal  angles  with  any  straight  line  intersecting  them. 


Fig.  it. 


Let  AC  and  BD  represent  the  two  lines,  BA  the  third 
line,  and  MO  any  other  line ;  and  let  the  A  CAK  and 
DBK  be  the  angles,  that  by  hypothesis  are  equal. 


\ 


20 


ELEMENTS   OF   GEOMETRY. 


/.KAO+/.  AKO  +Z  KOA  =180° 

Z  KBM+Z  BKM+Z  70/i?  =180° 

0       +       0 


Ex.  1.  §  16. 

Ex.  1.  §  16. 

+Z  KOA-Z  KMB=0,  by  subtraxjtion. 
.-.  Z  KOA  =  Z  /OfB,        Axiom  (c)  §  12. 

Q.E.  D. 

Ezercises.  —  1.  Establish  the  theorem  when  the  fourth  line 
passes  through  B. 


Proof.  ZBAO  +  ZAOB-h  Z  ABO  =  180°        Ex.  1,  §  16. 

ZABD-\-  ZDBH+  ZHBK=  180°  §11. 

0  +  Z  AOB  -  Z  DBH  +0=0 
By  subtraction,  ._  Z  AOB  =  Z  DBH  Axiom  (c),  §  12. 

Q.  E.D. 

2.  Establish  the  theorem  when  the  fourth  line  does  not  inter- 
sect the  third  within  the  limits  of  the  drawing. 

Hint.  —  Draw  an  auxiliary  line  connecting  two  points  of  inter- 
section and  then  apply  Ex.  1.     Two  applications  will  be  necessary. 

S.  Establish  the  theorem  when  the  third  and  fourth  lines 
intersect  between  the  first  and  second. 

Note. — In  order  to  thoroughly  familiarize  himself  with  the 
methods  of  procedure  and  with  the  results,  the  student  should 
make  for  himself  figures  that  differ  from  those  in  the  text  both 
in  relative  proportions  and  in  lettering,  and  use  them  to  make 
demonstrations.  He  should  be  accurate  and  complete  in  enunci- 
ation and  in  demonstration,  quoting  his  authority  for  each  state- 
ment. Remember  that  neatness  and  accuracy  of  expression, 
whether  written  or  oral,  tend  to  accuracy  of  thought. 


PARALLELS.  21 


PARALLELS. 

18.  Definition.  If  two  straight  lines  in  a  plane  make 
equal  angles  with  a  third  straight  line  in  the  plane,  the 
two  are  said  to  be  parallel  (II). 

The  third  is  called  a  secant,  or  sometimes  a  transversal. 

By  §  17,  parallels  make  equal  angles  with  any  secant. 

AC  and  BD  are  said  to  have  the  same  direction. 
Their  opposites  AF  and  BG  have  the  same  direction. 


D 


Fig.  19. 

Exercises.  —  1.  Show  that  the  eight  angles  about  A  and  B  are 
in  two  sets  ;  those  in  one  set  being  equal  to  each  other,  and  those 
in  the  other  set  being  equal  to  each  other. 

2.  Show  that  the  A  CAB  and  DBA  are  supplementary.  Show 
the  same  of  FAB  and  ABG. 

19.  Theorems,  (a)  If  the  two  lines  which  form  an 
angle  are  parallel  to  the  two  lines  which  form  another  angle, 
the  angles  icill  be  equal,  if  the  lines  forming  one  angle 
extend  from  its  vertex  in  the  same  direction  as  the  lines 
forming  the  other  angle. 

(b)  If  the  lines  forming  one  angle  extend  from  its  vertex 
in  the  opposite  direction  to  the  lines  forming  the  other  angle, 
the  angles  ivill  be  equal. 


22 


ELEMENTS   OF   GEOMETEY. 


(c)  If  one  set  of  parallels  extend  in  the  same  direction, 
and  the  other  set  in  opposite  directions  from  the  vertices,  the 
angles  will  be  supplementary. 


Fig.  20. 

Let  A  and  B  represent  the  two  vertices.  The  student 
will  supply  the  demonstrations  called  for  by  these 
theorems.     The  figure  furnishes  suggestions. 

Exercises.  —  1.  Show  that  if  a  line  is  perpendicular  to  one  of 
two  parallels,  it  is  perpendicular  to  the  other  also. 

2.  Show  that  parallels  are  everywhere  equally  distant  from 
each  other. 


A 

m\ 

B 

D 

N\ 

C 

Fig.  21. 


Proof.  —  If  the  segments  AD  and  BC  (which  measure  the  dis- 
tances between  the  parallels  AB  and  DC  aX  A  and  B)  are  equal, 
it  is  probable  that  they  may  be  brought  to  coincide  and  their 
equality  be  thus  established. 


PARALLELS.  23 

If  at  Ml  the  middle  point  of  the  segment  AB,  an  auxiliary  line 
MN  be  drawn  perpendicular  to  AB  (§  13)  it  will  be  perpendicular 
to  Z>C  (§  19,  Ex.  1).  But  we  do  not  as  yet  know  whether  it  will 
bisect  DC  OT  not. 

If  we  revolve  the  portion  of  the  figure  that  lies  on  the  right  of 
the  line  MN  (§§  8  and  10)  about  MN  as  an  axis  until  the  revolved 
portion  coincides  with  the  unrevolved  portion,  we  shall  have  the 
segment  MB  coinciding  with  the  segment  3IA  (because  M  is  by 
selection  the  middle  point  of  the  segment  AB).  The  poipt  B  will 
fall  at  A,  and  the  line  BC  (being  perpendicular  to  MB)  will  in  its 
revolved  position  take  the  direction  AD. 

Because  NC  is  perpendicular  to  MN,  it  will  when  rotated  lie  in 
the  direction  ND. 

The  point  C  will  then  lie  somewhere  in  the  line  AD,  and  some- 
where in  the  line  ND.  It  must  lie  at  their  intersection  (D). 
Therefore  the  segment  BC  will  coincide  with  the  segment  AD  and 
be  equal  to  it.  Q.  e.  d. 

3.  Show  that  if  two  lines  are  parallel  to  a  third  line,  they  are 
parallel  to  each  other. 

4.  Show  how  perpendiculars  to  a  line  may  be  drawn  from 
points  on  the  line.  Let  fall  perpendiculars  from  given  points  to  a 
given  line.     Through  given  points  draw  parallels  to  a  given  line. 

Suggestion.  —  Use  a  ruler  and  a  right-angled  triangle ;  and  as- 
sume the  line  in  a  variety  of  positions. 

Note.  —  The  formal  statement  of  a  theorem  may  be  followed  by 
the  proof ;  or  the  I'elatious  leading  to  conclusions  may  be  presented 
first  and  the  formal  statement  of  the  theorem  be  presented  at  the 
end.  When  placed  before  the  proof,  it  is  a  statement  of  relations 
said  to  exist.  The  proof  is  the  establishing  of  these  said  relations 
(see  §  17).  Wlien  placed  after  the  determination  of  relations  it  is 
in  the  form  of  a  conclusion  (see  §  16). 

In  general  it  is  better  to  state  the  theorem  first,  so  that  the 
student  shall  have  in  mind  the  relation  that  he  is  undertaking  to 
establish. 


24 


ELEMENTS   OF   GEOMETRY. 


20.  Theorem.  If  at  the  middle  point  of  a  segment  of  a 
line  a  perpendicular  be  erected,  and  if  from  any  point  in 
the  perpendicular  lines  be  draicn  terminating  at  the  ex- 
tremities of  the  segment,  they  mil  be  equal  to  each  other. 


Fig.  22. 

If  PB  =  PA,  it  can  be  brought  to  coincide  with  PA, 
and  since  AB  and  MP  are  perpendicular  to  each  other,  a 
rotation  of  one  part  of  the  figure  on  a  line  as  an  axis  is 
suggested. 

Proof  Eevolve  the  portion  of  the  figure  that  lies  to  the 
right  of  the  line  MP,  on  MP  as  an  axis,  until  it  coincides 
with  the  plane  on  the  left  of  MP.  §  10. 

The  Z  PMB  will  coincide  with  the  Z  PMA.  B  will 
fall  aX.  A;  P  will  remain  stationary;  and  the  segment 
PB  will  coincide  with  the  segment  PA,  and  must  there- 
fore be  equal  to  it.  q.  e.  d. 

Ezercises.  —  1.  Show  that  any  point  not  on  the  perpendicular 
bisector  will  not  be  equally  distant  from  the  extremities  of  the 
bisected  segment. 

2.  Show  that  if  a  line  have  two  of  its  points  equally  distant  from 
the  ends  of  the  segment,  it  will  be  the  ±  bisector  of  the  segment. 


A  CIRCLE.  25 

Solution.  —  If  a  perpendicular  bisector  of  the  segment  were 
erected  it  would  contain  all  points  that  are  equally  distant  from 
the  extremities  of  the  segment.  For  this  reason  it  would  pass 
through  the  two  points  mentioned  in  the  hypothesis;  but  '■^two 
points  determine  the  position  of  a  straight  line." 

Therefore  the  line  which  by  hypothesis  passed  through  two 
points  that  were  equally  distant  from  the  extremities  of  a  given 
segment  will  coincide  with  and  will  be  the  perpendicular  bisector 
of  the  given  segment.  q.  e.  d. 

Note.  —  The  perpendicular  bisector  of  the  segment  of  a  line  is 
said  to  be  the  loctis  (place)  of  the  point,  when  moving  so  that  its 
distances  from  the  segment  ends  shall  always  be  equal  to  each  other. 

Or  it  may  be  described  as  the  locns  of  a  point  moving  so  that  the 
ratio  of  its  distances  from  two  fixed  points  always  equals  unity, 

A  CIRCLE, 

21.  Definitions.  The  locus  of  a  moving  point,  the  dis- 
tance of  which  from  a  given  point  is  fixed,  is  called  a 
circumference. 


Fig.  23. 


If  the  line  AB  rotate  about  ^  as  a  pivot,  any  point  in 
the  line  AB,  as  the  point  B,  in  a  complete  rotation,  will 
remain  at  a  fixed  distance  from  A,  and  will  generate  a 
circumference. 


26  ELEMENTS   OF   GEOMETRY. 

Anything  less  than  a  complete  rotation  will  generate 
an  arc.  A  half  rotation  will  generate  a  semi-circumfer- 
ence ;  and  a  quarter  rotation  will  generate  a  quadrant. 

The  point  A  is  called  the  centre ;  and  the  distance  of 
B  from  A  is  called  the  radius. 

A  point  nearer  to  A  than  B  is,  Mali  generate  a  circum- 
ference that  will  lie  entirely  within  the  circumference 
generated  by  the  point  B;  and  a  point  at  a  greater  distance 
from  A  than  B  is,  will  generate  a  circumference,  lying 
entirely  outside  of  the  circumference  generated  by  B. 

When  AB  rotates  about  A  as  a.  pivot,  the  change  of 
direction  makes  an  angle  at  A ;  each  point  of  AB  gener- 
ates a  circumference,  and  the  whole  line  generates  the 
surface  of  the  plane. 

The  figure  bounded  by  a  circumference  is  called  a  circle. 

CONGRUENT  TRIANGLES. 

22.  Definition.  When  a  geometric  figure  may  be  sub- 
stituted for  another  and  may  be  made  to  occupy  exactly 


Fig.  24. 


the  same  space  which  the  other  did,  the  figures  are  said 
to  be  congruent 


TRIANGLES. 


27 


Theorem.  If  tioo  triangles  have  the  three  sides  of  the 
one  equal  to  the  three  sides  of  the  other,  each  to  each,  they 
are  congruent. 

If  the  two  triangles  {p)  and  (g)  having : 
AB  =  DE, 
BC=DF, 
and  AC=EF, 

be  so  placed  that  a  pair  of  equal  sides  shall  lie  together, 
and  the  triangles  be  not  superimposed,  we  shall  have 
them  placed  as  in  figure  (r). 

By  the  terms  of  the  theo- 
rem, B  and  C  are  two  points 
equally  distant  from  A  and  E. 

Draw  the  auxiliary  line 
AE. 

By  §  20,  Ex.  2,  the  line 
CB  will  be  the  perpendicular 
bisector  of  the  segment  AE. 

If  the  figures  to  the  right 
of  the  line  CK  be  revolved  on  CK  as  an  axis,  the  point 
E  will  fall  at  A ;  the  segments  ED  and  AB  will  coin- 
cide; the  same  will  be  true  of  ^i^and  AC;  DF  and  BC 
will  remain  in  coincidence;  and  all  the  parts  of  one 
triangle  (perimeter,  angles,  and  surface)  will  coincide 
with  the  parts  of  the  other.  q.  e.  d. 

23.  Theorem.  If  two  triangles  have  tico  angles  and 
an  included  side  of  one  equal  to  tioo  angles  and  an  included 
side  of  the  other,  they  are  congruent. 

he  student  will  make  the  necessary  constructions,  and 
show  that  the  triangles  may  be  placed  so  as  to  coincide. 


Fig.  25. 


mae 

shoA 


28 


ELEMENTS   OF   GEOMETRY. 


Exercise.  —  Show  that  two  triangles  which  have  any  two  angles 
and  a  side  of  one  equal  to  two  angles  and  the  corresponding  side 
of  the  other  are  congruent. 

Note.  —  Corresponding  side  means  one  that  is  placed  in  like 
relation  to  the  equal  angles. 

Corresponding  sides  are  often  called  homologous  sides. 

24.  Theorem.  If  two  triangles  have  two  sides  and  the 
included  angle  of  one  equal  to  two  sides  and  the  included 
angle  of  the  other,  they  are  congruent. 

The  student  will  supply  the  necessary  constructions, 
and  make  application  of  one  figure  to  the  other. 

Exercise.  —  Show  what  significant  position  two  triangles  may 
be  made  to  occupy  with  respect  to  each  other,  when  two  sides  of 
one  are  equal  to  two  sides  of  the  other  and  the  included  angles  are 
supplementary. 

CONSTRUCTION  OF  TRIANGLES. 
Four  Cases. 

25.  First  Case.     When  three  sides  are  given. 


« 


Fio.  26. 


On  the  line  MN  lay  off  a  segment  (a)  with  ruler,  taj 
or  compasses.     C  and  B  will  mark  two  of  the  vertices 


CONSTEUCTION    OF   TRIANGLES.  29 

i^equired  triangle.  With  (7  as  a  centre  and  a  radius 
(6)  construct  an  arc  ;  and  with  jB  as  a  centre  and  radius 
(c)  construct  another  arc.  From  the  intersection  A, 
draw  the  lines  AC  and  AB. 

Note.  —  It  is  customary,  as  a  matter  of  convenience,  to  desig- 
nate the  vertices  of  a  triangle  by  tlie  capital  letters,  and  the  sides 
opposite  to  them  by  the  same  letters  in  small  type. 

The  triangle,  the  vertices  of  which  are  A,  B,  and  C 
(A  ABC),  will  be  the  required  triangle.  q. e.  r. 

The  construction  can  be  made  when  the  sum  of  any 
two  of  the  segments,  given  as  sides,  is  greater  than  (>) 
the  third  side ;  but  not  otherwise. 

The  student  will  substantiate  this  statement. 

Exercises. — 1.  Show  that  if  two  circumferences  intersect  in 
one  point,  they  will  intersect  in  two  points. 

2.  Show  that  the  same  wiU  be  true  if  a  straight  line  intersects 
a  circumference. 

3.  Show  that  in  the  figure  the  A  CKB  is  congruent  with  the 
A  CAB. 

4.  Show  that  two  other  triangles  might  be  constructed  by  inter- 
changing the  radii  (b)  and  (c) ;  and  that  the  four  triangles  thus 
constructed  would  be  congruent. 


Fig.  27. 


5.  With  ruler  and  compass  show  how  to  construct,  in  any  posi- 
tion, an  angle  that  shall  be  equal  to  a  given  angle. 

Suggestion.  —  Let  0  (Fig.  p)  represent  the  given  angle.  If  we 
draw  any  line  (EF),  intersecting  tlje  lines  forming  the  angle,  so  as 


80  ELEMENTS   OF   GEOMETRY. 

to  obtain  a  triangle  having  d  for  one  of  its  angles,  we  shall  have  a 
triangle,  the  three  sides  of  which  may  be  used  to  construct  a  con- 
gruent triangle  in  any  desired  position,  as  in  (Fig.  q). 

It  is  customary,  as  a  matter  of  convenience,  to  assume  AF  and 
AE  equal  to  each  other. 

26.    Second  Case.     That  in  whicli  two  angles  and  the 
included  side  are  given. 


Fio.  28. 

On  the  indefinite  line  BC  lay  off  the  segment  BC  equal 
to  the  given  side,  and  at  B  and  C  construct  the  given 
angles,  so  that  they  shall  be  interior. 

There  will  be  two  constructions,  but  the  two  will  be 
congruent  and  may  be  considered  as  one. 

Note.  —  This  case  may  be  considered  as  including  the  one  in 
which  are  given  two  angles  and 
a  side  opposite  one  of  them  ;  for 
if  two  angles  are  given,  the  third 
may  be  determined,  and  the 
method  of  this  case  may  then  be 
applied.  Fio.  29. 

Exercises.  —  1.  Make  a  construction  when  a  =  3  inches, 
e  =  80°,  and  0  =  120°. 

2.  Show  how  to  construct  a  triangle,  if  we  have  given :  a  side, 
an  angle  adjacent,  and  an  angle  opposite. 


CONSTRUCTION   OF   TRIANGLES. 


31 


27.  Third  Case.  That  in  which  two  sides  and  the 
inchided  angle  are  given.  Construct  the  angle ;  from  the 
vertex  lay  off  the  given  sides ;  and  join  their  extremities. 


Fio.  80. 

The  student  will  show  that  there  are  two  constructions, 
but  the  resulting  triangles  are  congruent. 

28.  Fourth  Case.     That  in  which  two  sides  and  an 
angle  opposite  one  of  them  are  given. 

a C^ 


Construct  an  angle  equal  to  6.  On  either  side  of  the 
angle  lay  off  the  segment  (6).  Let  C  be  its  other  ex- 
tremity. From  C  as  a  centre  with  a  radius  (a)  construct 
an  arc.  Join  C  with  the  points  in  which  the  arc  inter- 
sects the  other  side  of  the  angle.     (See  §  25,  Ex.  2.) 

Exercises.  —  1.  Show  that  if  (a)  be  less  than  the  ±  (p),  there 
cannot  be  any  construction. 

2.  Show  that  if  a  =p,  there  can  be  one  construction,  provided 
0  <  90°,  and  not  any  if  ^  ^  90°. 

3.  Show  that  if  a  be  intermediate  in  value  between  b  and  p, 
(6  >  rt  >p),  there  can  be  tico  constructions  if  ^  <  90°,  and  none 
if  61  >  90°. 


32 


ELEMENTS   OF   GEOMETRY. 


4.  Show  that  if  a  =  b,  there  can  be  one  construction  when 
e  <  90°,  and  none  when  6  >  90°. 

6.  Show  that  if  a  >  6,  there  can  be  one  construction  when 
e  <  90°,  and  one  when  6  >  90°. 

6.  Show  that  if  the  three  angles  are  given,  the  triangle  will  not 
be  determined. 

Notes.  — 1.  Wlien  two  or  more  constructions  (not  congruent) 
are  possible,  the  case  is  said  to  be  ambiguoiLs. 

2.  In  order  to  construct  a  plane  triangle,  three  parts  must  be 
given,  one  of  which  is  a  line. 


Fig.  82. 


GENERAL  EXERCISES. 

1.  Show  that  lines  which  are  not  parallel  will  intersect. 

2.  Show  how  to  bisect  a 
given  segment  of  a  line. 

8.  Show  that  oblique  lines, 
from  a  point  in  a  perpendicular, 
intersecting  the  base  line  at 
different  distances  from  the 
foot  of  the  perpendicular,  make 
unequal  angles  with  the  base 
line. 

4.  Show  that  an  exterior 
angle  of  a  triangle  equals  the 
sum  of  the  interior  non-adja- 
cent angles. 

Inequality  Axioms. 
(a)  If  the  greater  members  of 
two  inequalities  he  added,  the 
Slim  will  he  greater  than  the 
sum  of  the  lesser  members. 

(b)  If  equals  he  added  to  or  subtracted  from  the  ttvo 
members  of  an  inequality,  the  inequality  will  still  exist,  and 
will  exist  in  the  same  sense. 


Fig.  88. 


CONSTRUCTION   OF   TRIANGLES. 


33 


(c)  If  the  members  of  an  inequality  he  multiplied  or  he 
divided  hy  a  positive  quantity,  the  inequality  icill  suhsist  in 
the  same  sense;  hut  if  midtiplied  or  divided  hy  a  negative 
quantity,  the  inequality  will  he  reversed. 

6.  Show  that  the  sum  of 
two  sides  of  a  ti-iangle  is  greater 
than  the  sum  of  the  distances 
from  any  point  within  the  tri- 
angle to  the  extremities  of  the 
third  side. 

a  +  6  >  c  +  d 
d+e>f 


a+b+e>c+f 

6.  Show  that  if  from  any 
I)oint  in  a  perpendicular,  oblique 
lines  be  drawn  to  the  base  of 
any  two,  the  one  which  meets 
the  base  at  the  greater  distance 
from  the  foot  of  the  perpendicu- 
lar will  be  the  greater.  Fig.  36. 

7.  Show  how  to  construct  a  triangle,  having  given  two  angles 
and  the  side  opposite  one  of  them  without  having  found  the  third 

angle. 

8.  Show  that  the  difierence  of  two  sides  of  a  triangle  is  less 
than  the  third  side. 

9.  Through  a  point  to  draw  a  line  parallel  to  a  given  line,  and 
show  that  only  one  such  line  can  be  drawn. 

10.  Show  how  we  may,  with  ruler  and  compass,  erect  a  perpen- 
dicular at  any  point  of  a  line,  and  how  we  may  let  fall  a  pei-pendic- 
ular  from  a  point. 

11.  Prove  the  converse  of  Ex.  2,  §  18. 

Note.  — A  proposition  is  a  statement  of  a  relation  said  to  exist. 
and  is  in  the  general  form  of  subject  and  predicate. 

D 


84  ELEMENTS   OF  GEOMETRY. 

The  converse  of  a  proposition  is  also  a  proposition  ;  but  with 
relations  of  subject  and  predicate  reversed. 

Exercise  2,  §  18,  as  a  formal  proposition  would  be  :  (If  the  two 
lines  AC  and  BD  are  parallel  and  are  intersected  by  a  third  line 
HE)(then  will) (the  angles  BAC  and  ABD  be  supplementary). 
The  converse  would  be  :  (If  two  lines  are  met  by  a  third  line  so  as 
to  make  the  interior  angles  on  one  side  of  the  secant  supple- 
mentary) (then  will)  (the  two  lines  so  situated  with  rfespect  to  the 
third  be  parallel). 

The  Beductio  ad  Absurdttm  method  is  particularly  well  adapted 
to  the  determination  of  the  truth  or  falsity  of  the  converse  of  a 
proposition. 


CHAPTER    III. 

29.    Definitions.     A  triangle  is  called : 

Eight,  when  one  of  its  angles  is  90°. 

Oblique,  when  none  of  its  angles  are  90°. 

Obtuse,  when  one  of  its  angles  is  >  90°. 

Acute,  when  each  of  its  angles  is  <  90°. 

Equiangular,  when  the  three  angles  are  equal  to  each 
other. 

Equilateral,  when  the  three  sides  are  equal  to  each  other. 

Isosceles,  when  two  sides  are  equal  to  each  other  and 
not  equal  to  the  third  side. 

Scalene,  when  there  is  not  any  equality  between  sides. 


F1S.-86. 

Special  names  given  to  parts  of  triangles  are : 

Base,  Avhich  may  be  any  one  side. 

Base  angles,  Avhich  are  the  angles  adjacent  to  the  base. 

Vertex,  which  is  the  point  of  intersection  of  the  sides, 
not  considered  the  base. 

Vertex  angle,  which  is  the  angle  of  the  triangle  at  the 
vertex. 

36 


86 


ELEMENTS   OF   GEOMETRY. 


Hypothenuse,  which  is  the  side  opposite  a  right  angle. 
Altitude,  which  is  the   perpendicular   from  vertex   to 
base. 

30.  Theorem.  If  aJt  the  middle  of  a  side  of  an  equi- 
lateral triangle  a  perpendicular  he  erected,  it  will  pass 
through  the  opposite  vertex. 

If  at  M,  the  middle  point  of  the  side  AB,  a  perpendicu- 
lar be  erected,  it  will  contain 
all  points  that   are  equally 
distant  from  A  and  B.     C  is 
such  a  point.  q.  e.  d. 

Ezercisea.  —  1.  Show  that  the 
angle  at  the  vertex  will  be  bi- 
sected. 

2.  Show  that  if  perpendiculars 
be  let  fall  from  the  three  vertices 
to  the  opposite  sides  of  an  equi- 
lateral triangle,  they  will  be  equal 
to  each  other. 

8.   Establish  the  converse  of  the  fio.  8T. 

theorem. 

4.  Show  that  an  equilateral  triangle  is  equiangular. 
6.  Establish  the  converse. 

31.  Theorem.  If  a  perpendiaidar  he  erected  at  the 
middle  point  of  the  non-equal  side  of  an  isosceles  triangle, 
it  will  pass  throxigh  the  opposite  vertex. 

The  proof  is  the  same  as  in  §  30. 

Ezercises.  —  1.  Show  that  the  angles  opposite  the  equal  sides 
are  equal. 

5.  Show  that  the  angle  at  the  vertex  is  bisected  by  the 
perpendicular. 


TRIANGLES.  37 

3.   Show  that  if  perpendiculars  be  let  fall  from  the  vertices  of 
the  equal  angles,  they  will  be  equal  to  each  other. 


Solution.  —  Let  ACB  represent  the  isosceles  triangle  ;  and  AP 
and  BR  the  perpendiculars. 

If  the  perpendiculars  are  equal,  the  k^APC  and  BBC  will  be 
equal  because  they  will  then  have  a  pei'pendicular  and  an  hypoth- 
enuse  of  a  right  triangle  the  same  in  each. 

Does  this  relation  (the  equality  of  the  triangles)  exist  without 
the  consideration  of  the  perpendicular  ? 

In  the  A  APC  and  BE  C, 

AC  =  BC,  (by  hypothesis) 

^APC  =  ZBBC,  (being  90^) 

ZACP=^ZBCB,  (being  vertical) 

Hence  by  (§23,  Ex.),  A  APC  =  ABPC. 
AP&nd  BB  are  corresponding  parts  and  are  equal. 

The  question  has  thus  been  answered :  The  necessary  relation 
does  exist,  and  the  problem  is  solved.  q.e.d. 

4.  Solve  the  same  problem,  using  for  the  figure  an  acute- 
angled  triangle. 

5.  Show  that  if  two  angles  of  a  triangle  are  equal,  the  sides 
opposite  those  angles  are  equal,  i.e.  the  triangle  wUl  be  isosceles. 

Note.  —  A  problem  is  something  proposed  to  be  done.  The 
solution  is  the  finding  of  sufiicient  previously  established  relations 
to  warrant  the  doing. 


111116 


38 


ELEMENTS  OP  GEOMETRY. 


32.  Theorem.  If  two  angles  of  a  triangle  are  unequal, 
the  sides  opposite  them  are  unequal,  and  the  greater  side  lies 
opposite  the  greater  angle. 

Let  ABC  represent  the  triangle,  the  angles  of  which  at 
A  and  B  are  unequal. 

/jf  at  the  middle  point  {M) 
of  the  side  AB  a  perpen- 
dicular be  erected,  it  could 
not  pass  through  C;  for  if 
it  did,  the  triangle  would 
be  isosceles.  Not  passing 
through  C,  it  must  intersect 
the  lines  forming  the  other 
sides  of  the  triangle  at  sepa- 
rate points.  Let  D  be  the  intersection  that  is  the 
nearer  to  M. 

Draw  the  auxiliary  line  DB.  It  will  separate  the  angle 
to  which  it  is  drawn  into  two  parts,  one  of  which  equals 
the  Z  CAB.    The  Z  5  is  the  larger. 

AC  =  AD -{- DC  =  BD -\- DO  BC.       q.e.d. 


Fig.  89. 


Exercises.  — 1.  Prove  the  same  by  using  AE  as  an  auxiliary 
line. 

2.  Show  that  the  hypothenuse  of  a  right  triangle  is  greater  than 
either  of  the  other  sides. 

3.  Establish  the  theorem  of  this  section,  with  an  obtuse-angled 
triangle. 

4.  Establish  the  converse  of  the  theorem. 

6.  Show  that  if  two  right  triangles  have  one  of  the  sides  adja- 
cent to  the  right  angle  and  the  hypothenuse  mutually  equal,  they 
are  congruent. 


TRIANGLES. 


39 


33.    Theorem.    Any  point  of  an  angle  bisector  is  equally 
distant  from  the  lines  forming  the  angle. 


Fig.  40. 


If  AP  represent  the  angle  bisector,  and  if  from  any 
point  (P)  perpendiculars  be  let  fall  to  the  lines  forming 
the  angle,  the  A  PAF  and  PAB  will  be  congruent,  having 
two  angles  and  the  included  side  of  one  equal  to  two 
angles  and  the  included  side  of  the  other. 

Hence  PF  =  PB.  q.  e.  d. 

Theorem.  TTie  distances  of  any  point,  not  on  the  angle 
bisector,  from  the  angle  lines  will  not  be  eqvxjX. 


Fig.  41. 


Let  Q  represent  any  point  not  on  the  angle  bisector. 
Let  QD  and  QK  be  the  perpendiculars.     One  of  them 


40 


ELEMENTS   OF   GEOMETRY. 


intersects  the  angle  bisector.     Let  U  be  the  point  of 
such  intersection. 
Draw  EC  ±  to  Aff",  and  join  QC. 

QD  =  QH-i- HD  =  QH+  HC>QC>  QK 

..QD>QK.  Q.E.ii. 

Exercise.  —  Show  that  the  bisector  of  the  supplement  of  Z  KAD 
will  be  perpendicular  to  AH,  and  that  every  point  in  it  will  be 
equidistant  from  the  angle  lines. 

Note.  —  If  perpendiculars  on  one  side  of  a  line  are  positive 
(+),  those  on  the  other  side  are  negative  (  — )• 

The  locus  of  a  point,  the  ratio  of  the  distances  of  which  from 
two  lines  is  (+  1),  will  be  the  angle  bisector  If  the  ratio  of  the 
distances  is  (—1),  the  locus  will  be  the  bisector  of  the  supple- 
ment of  the  first  angle  and  will  be  perpendicular  to  the  bisector  of 
the  first  angle. 

34.   Problem.     To  construct  a  bisector  of  an  angle. 


Fio.  43. 


If  AP  were  the  angle  bisector  required,  and  if  we 
should  draw  the  perpendiculars  PB  and  PF,  they  would 
be  equal  to  each  other,  and  the  segments  AB  and  AF 
woiUd  be  equal.  If  the  auxiliary  line  BF  were  drawn, 
AP  would  be  a  perpendicular  bisector  to  it. 


TRIANGLES. 


41 


This  determination  of  relations  that  would  exist  if  the 
angle  bisector  were  drawn  suggests  the  construction. 


Fig.  48. 


Lay  off  AB  equal  to  AF,  and  find  some  point  (K), 
other  than  (A),  which  is  equally  distant  from  B  and  F. 

Draw  AJ{^;  it  will  be  the  angle  bisector;  since  it  is  the 
perpendicular  bisector  of  BF. 

Note. —  We  might  have  followed  more  closely  the  relations  at 
first  developed,  by  erecting  at  B  and  F  perpendiculars  to  AB  and 
AF,  and  have  joined  their  intersection  with  A. 


Fig.  44. 


35.  If  through  the  point  P  a  perpendicular  to  the  line 
AB  be  dra^vn,  and  then  be  rotated  positively  (i.e.  to  the 
left),  the  point  in  which  it  intersects  AB  will  move  from 
F  in  the  direction  of  B.  When  it  shall  have  rotated  90°, 
the  line  through  P  will  be  parallel  to  AB  and  will  not 
intersect  it.  We  mean  the  same  thing  when  we  say  that 
the  point  of  intersection  has  passed  to  infinity. 


42 


ELEMENTS   OF   GEOMETRY. 


The  instant  the  angle  exceeds  90°,  the  rotating  line 
again  intersects  the  line  AB,  in  the  direction  FA.  The 
distance  from  F  will  continually  diminish  until  180°  of 
rotation  has  taken  place,  when  the  point  of  intersection 
^v^ll  have  reached  F. 

The  next  180°  the  point  of  intersection  will  travel  over 
the  same  route  as  before. 


QUADRILATERALS  AND   QUADRANGLES. 

36.   A  plane  figure  formed  by  four  straight  lines  which 
enclose  an  area  forms  a  quadrilateraL     If  each  line  inter- 


FlO.  45.  —  QtTADBILATUiAL. 

sects  each  of  the  others  as  indicated  in  the  above  figure, 
the  figure  is  called  a  complete  quadrilateral. 

The  portion  ABCD  is  called  a  quadrangle,  the  vertices 
of  which  are  -at  A,  B,  C,  and  D,  and  the  sides  of  which 
are  AB,  BC,  CD,  and  DA. 


Fio.  4«.  —  TRAPKZorD. 


-  Paralleix>oram. 

If  two  sides  are  parallel,  the  quadrangle  is  called  a 
trapezoid. 


QUADRILATERALS   AND   QUADRANGLES. 


43 


If  the  sides  of  the  quadrangle  are  parallel  two  and  two, 
the  figure  is  called  a  parallelogram. 


Fig.  48. — Bectangle. 


Fig.  49.  —  Sqtjaeb. 


If  a  parallelogram  be  right-angled,  the  figure  is  called 
a  rectangle. 

If  the  rectangle  have  all 
its  sides  equal  to  each  other, 
the  figure  is  called  a  square. 

If  the  sides  of  an  oblique 
parallelogram  (none  of  the 
angles  being  90°)  are  equal, 
the  figure  is  called  a  rhombus.  A  quadrangle  not  having 
any  side  parallel  to  any  other  side  is  sometimes  called 
a  trapezium. 

37.    Theorem,     The  sum  of  the  exterior  angles  of  a 
quadrangle  is  360°,  or  four  right  angles. 


Fig.  50.  —  Rhombtts. 


Fig.  51. 


44  ELEMENTS   OF   GEOMETRY. 

A  point  S  moving  in  the  direction  indicated,  about  the 
perimeter  of  the  quadrangle  and  arriving  at  its  initial 
position,  will  have  made  lejl-hsinded.  changes  of  direction 
at  the  four  vertices.  These  changes  of  direction  being 
placed  adjacent  to  each  other,  as  indicated  in  the  second 
figure,  give  a  complete  rotation,  or  360°.  q.  e.  d. 

Exercise.  —  Show  that  the  sum  of  the  interior  angles  is  four 
right  angles. 

38.  Theorem.  The  opposite  sides  of  a  parallelogram 
are  equal. 

If  the  theorem  be  true,  and  we  draw  an  auxiliary  line 
connecting  a  pair  of  non-adjacent  vertices  (such  a  line 
is  called  a  diagonal),  we  would  have  the  figure  separated 


Fia.  52. 

into  two  triangles,  which  would  have  the  three  sides  of 
one  equal  to  the  three  sides  of  the  other,  and  would  be 
equal. 

Can  we  show  that  the  triangles  are  equal  (congruent), 
making  use  only  of  previously  established  relations  ? 
We  can,  because : 

Z  CBD  =  Z  ADB, 
Z.  CDB  =  Z  ABD, 
and  BD  =  BD. 


QUADRILATERALS   AXD    QUADRANGLES.  45 

The  two  triangles  have  two  angles  and  the  included 
side  in  each  equal,  and  are  therefore  congruent  (§  23) ; 
and  hence  the  theorem. 

Exercises.  —  1.  Use  the  theorem  to  show  that  parallels  are 
everywhere  equally  distant  from 
each  other.  A jq 

2.  If  a  pair  of  non-adjacent 
sides  in  a  quadrangle  are  paral- 
lel   and   equal,   show  that    the 

figure  will   be  a   parallelogram    C  ~  T) 

(O). 


Fig.  53. 


39.    Theorem.     Tlie  diagonals  of  a  parallelogram  mut- 
ually bisect  each  other. 


Fig.  M. 

If  the  diagonals  do  bisect  each  other,  the  A  BIC  and 
AID  will  be  equal. 
Are  they  ? 
They  are,  because : 

Z  IBC  =  Z  IDA, 
Z  ICB  =  Z  IAD, 
and  BC  =  AD. 

From  which  A  BIC  =  A  AID. 

.'.  BI=  ID  and  AI=  IC. 

Q.  E.  D. 


46  ELEMENTS   OF   GEOMETRY. 

Exercises.  —  1.  Show  that  the  diagonals  of  a  rectangle  are 
equal  to  each  other. 

2.  Show  that  the  diagonals  of  a  rhombus  are  perpendicular  to 
each  other. 

8.  Show  that  the  diagonals  of  a  square  are  equal,  and  are  per- 
pendicular to  each  other. 

40.  Definitions.  A  polygon  is 'a  figure  formed  by  a  num- 
ber of  straight  lines  which  enclose  an  area. 


Fio.65. 

In  a  polygon  there  are  as  many  angles  as  sides. 

A  Triangle      has  3  angles.  A  Nonagon      has   9  angles. 

A  Quadrangle  has  4  angles.  A  Decagon       has  10  angles. 

A  Pentagon     has  5  angles.  An  Undecagon  has  11  angles. 

A  Hexagon      has  6  angles.  A  Dodecagon   has  12  angles. 

A  Heptagon     has  7  angles.  A  Pendecagon  has  15  angles. 

An  Octagon     has  8  angles.  Etc. 

If  all  the  angles  are  equal  to  each  other,  and  all  the 
sides  are  equal  to  each  other,  the  polygon  is  regular. 

An  exterior  angle  of  a  polygon  is  the  change  of  direc- 
tion in  going  from  one  side  to  an  adjacent  side. 

Any  polygon  in  which  all  the  changes  of  direction  are 
in  the  same  sense,  is  called  a  convex  polygon. 

If  the  changes  of  direction  are  not  all  in  the  same 
sense,  the  polygon  is  said  to  be  re-entrant. 


POLYGONS. 


47 


41.   Theorem.     Tlie  sum  of  the  exterior  angles  of  any 
2?olygou  is  360°. 

If  on  the  perimeter  of  any  convex  polygon  as  repre- 
sented in  the  accompanying 
figure  we  take  any  point 
as  (S),  and  traverse  the 
perimeter,  starting  in  the 
direction  indicated,  and  re- 
tvirning  to  (S),  we  shall 
at  the  vertices.  A,  B,  C,  D, 
and  E,  have  made  changes 
of  direction  to  the  left, 
amounting  in  all  to  a  complete  rotation,  or  360°. 

Figure  (b)  represents  a  re-entrant  polygon. 

As  in  the  convex  polygon,  traversing  the  perimeter 
from  (S),  starting  in  the  direction  indicated  and  arriving 


Fig.  57. 

at  (S),  a  complete  rotation  will  have  been  made.  But  it 
is  to  be  noted  that  at  A  the  change  of  direction  is  to  the 
leji,  at  B  it  is  to  the  right,  at  C,  D,  E,  and  2^  it  is  to  the  left. 
Let  P  (Fig.  c)  be  any  point  in  the  plane.  From  P 
draw  lines  parallel  to  the  lines  of  Fig.  6,  which  indicate 
the  changes  of  direction  at  the  succeeding  vertices. 


48  ELEMENTS   OF   GEOlVrETRY. 

We  see  that  the  angles  at  A,  C,  D,  E,  and  F  are  jwsi- 
tive,  while  the  angle  at  B  is  negative.  And  we  see  that 
the  aggregate,  or  the  algebraic  sum,  is  360°.  q.  e.  d. 

Note.  — The  proof  is  not  in  any  way  affected  by  the  number  of 
sides  of  the  polygon  or  by  the  number  of  re-entrant  angles.  We 
are  then  entitled  to  draw  the  conclusion  we  have. 

An  interior  angle  has  been  defined  as  the  supplement  of  the 
corresponding  exterior  angle.  An  exterior  angle  has  been  defined 
as  the  change  of  direction  at  a  vertex.  Then  the  interior  angle  at  B 
would  be  180  —  (  —  ^)  =  180°  +tf ;  an  appropriate  value,  as  may  be 
seen  by  drawing  auxiliary  lines  from  B  to  D,  E,  and  F;  and  then 
taking  the  sum  of  the  interior  angles  of  the  triangles  thus  formed. 

Exercises.  —  1.  Find  the  sum  of  the  interior  angles  of  a  penta- 
gon.   Find  what  each  one  will  be  if  the  pentagon  be  regular. 
2.  Do  the  same  for  polygons  of  6,  7,  8,  9,  10,  11,  and  12  sides. 

ANALYSIS. 
The  Way  to  attack  a  Problem. 

42.  In  every  field  of  investigation  problems  are  pre- 
sented. These  problems  require  solution.  The  solving  of  a 
problem  is  the  determination  of  sufficient,  previously  estab- 
lished relations,  to  warrant  the  conclusion  which  is  said  to 
exist,  or  which  appears  to  exist,  or  which  is  desired  to  exist. 

The  manner  of  approaching  the  solution  of  any  problem 
is  the  same  in  all  subjects,  i.e.  we  are  to  approach  it 
through  the  analysis. 

When  one  makes  an  analysis,  he  asks  himself  one  of 
the  three  following  questions : 

{ said  \ 

If  tJie  relations  -I  ivhich  appear  y  to  exist,  do  exist,  what 
i  desired  ) 

are  the  necessary,  previously    established,   and    sufficient 
relations  ? 


EXERCISES, 


49 


If  the  necessary,  previously  established,  and  sufficient 
relations  are  found  and  applied,  the  problem  is  solved. 

Note.  —  Students  frequently  concern  themselves  with  problems 
which  they  have  no  business  to  attack,  for  the  reason  that  their 
information  is  not  suflBcient.  When  one  attempts  the  making  of 
the  analysis,  if  the  problem  be  an  inappropriate  one  for  him  to 
attempt,  he  will  presently  become  aware  of  that  fact ;  and  further 
time  need  not  be  wasted,  for  unless  sufficient  necessary  relations 
have  been  previously  established  by  him,  there- is  no  power  to  solve 
the  problem. 

The  most  important  thing  in  education  is  the  learning  to  make 
an  analysis  of  every  problem  which  we  desire  to  solve.  The  instant 
that  one  asks  himself  the  question  formulated  above,  he  puts  him- 
self in  the  proper  frame  of  mind  for  determining  whether  the 
stated^  apparent,  or  desired  relations  have  sufficient  foundation. 
It  emphasizes  the  necessity  of  having  facts  and  relations  developed 
in  their  proper  order,  each  with  sufficient  basis. 


GENERAL  EXERCISES. 

1.  If  two  angles  can  be  so  placed  as  to  have  a  common  side 
and  other  two  sides  intersect,  the  JJ 
sum  of  the  intersecting  sides  will  be 
greater  than  the  sum  of  the   non- 
intersecting  sides. 

2.  Show  that  if  from  any  point 
within  a  triangle,  lines  be  drawn  to 
two  vertices,  the  difference  of  the 
interior  angles  at  P  and  C  will  equal 
the  sum  of  the  A  PAG  and  PBC. 

3.  Show  that  if  two  triangles  have 
two  sides  of  one  equal  to  two  sides  of 
the  other,  and  their  included  angles 
unequal,  the  triangle  having  the 
greater  included  angle  will  have  the 
greater  third  side.  Fio.  59. 


50 


ELEMENTS   OF  GEOMETRY. 


Suggestion.  —  The  vertex  C  of  the  triangle  having  the  smaller 
included  angle  (Z  ABC<  A  ABD)  will,  when  a  pair  of  equal  sides 
are  placed  together,  fall  as  indicated  in  on^  of  the  figures. 


4.  Show  that  if  a  right  triangle  have  one  of  its  oblique  angles 
30°,  the  side  opposite  the  30°  angle  wUl  be  one-half  the  hypothe- 
nuse. 

6.  Show  which  of  the  eight  parts  of  a  quadrangle,  when  given, 
will  be  sufficient  to  determine  a  construction. 

6.  Show  that  the  bisectors  of  the  interior  angles  of  a  triangle  are 
concurrent  (pass  through  one  point). 

Bemark.  — Be  careful  not  to  assume  that  they  are  concurrent. 


CHAPTER   IV. 

Circles. 

43.  Definitions.  A  secant  of  a  curve  is  a  straight  line 
which  intersects  the  curve.  A  secant  will  intersect  a 
curve  in  two  points.  The  nature  of  a  curve  may  be  such 
that  a  secant  may  intersect  it  in  more  than  two  points. 

A  chord  is  the  segment  of  a  secant  between  two  points 
of  intersection.  The  portion  of  the  curve  between  the 
same  points  subtends  the  chord. 

The  centre  of  a  curve  is  a  point  through  which,  if  any 
straight  line  be  drawn  intersecting  the  curve,  the  chord 
will  be  bisected  at  that  point. 

A  diameter  is  a  chord  which  passes  through  the  centre. 

A  tangent  to  a  curve  is  a  straight  line  having  a  point 
in  common  with  the  curve,  and  having  the  same  direction 
that  the  generating  point  of  the  curve  has,  at  the  common 
point.     The  common  point  is  called  the  point  of  contact. 


Fig.  61. 


If  we  have  a  curve,  as  the  one  indicated  by  HABK, 
and  if  we  cause  any  secant,  as  AB,  to  rotate  about  A  as 
a  pivot,  so  that  B  shall  move  along  the  curve  toward 

51 


52  ELEMENTS   OF   GEOMETRY. 

A,  and  eventually  coincide  with  A,  the  secant  will  be  a 
secant  until  B  coincides  with  A,  when  it  will  be  a 
tangent ;  for  the  straight  line  will  at  that  instant  have 
the  same  direction  that  a  point  in  motion  along  the  curve 
will  have  at  A. 

A  tangent  is  sometimes  said  to  be  the  limit  toward 
which  the  secant  approaches  as  the  points  of  intersection 
approach  coincidence. 

A  point  being  position  and  not  having  magnitude,  B 
may  pass  through  and  beyond  the  position  A.  When 
that  happens,  the  rotating  line  will  have  again  become  a 
secant. 


Fio.  62. 


Note.  —  All  curves  have  secants  and  chords  ;  but  comparatively 
a  small  number  of  curves  have  centres  and  diameters.  The  circle 
18  one  of  this  small  number. 

44.  Theorem.  A  tangent  to  a  circumference  is  perpenr 
dicular  to  the  radius  drawn  to  the  point  of  cental. 

Analysis.  —  If  a,  tangent  to  a  circumference  be  perpen- 
dicular to  a  radius,  the  rotating  line  which  was  a  secant  and 
became  a  tangent  must  approach  an  angle  of  90°  with  the 
radius  to  the  point  of  rotation  and  arrive  at  90°  at  the  limit. 


CIRCLES. 


63 


Proof.  —  Let  AB  be  a  tangent  at  A.  Draw  any  secant 
as  AJ^I.  The  A  ACM  is  isosceles,  and  the  Z  CAM  is  less 
than  90°.  As  the  secant  rotates  about  ^  as  a  pivot,  and  M 
approaches  A,  the  Z  ACM  will  approach  0,  and  the 
Z  CAM  wiU  approach  90°. 


Fig.  63. 


After  M  has  passed  through  A,  the  angle  CAM" 
will  be  greater  than  90°,  because  Z  GAK  is  less  than 
90°. 

As  the  change  has  been  a  continuous  one,  the  angle 
made  with  CA  changing  gradually  from  an  acute  to  an 
obtuse  angle  must  have  passed  through  90°.  And 
furthermore,  when  the  second  point  is  on  either  side  of  A 
the  angle  is  oblique,  it  must  be  90°  when  the  second 
point  coincides  with  the  pivotal  point,  and  the  rotating 
line  has  become  a  tangent,  q.  e.  d. 

45.  Theorem.  The  perpendicidar  bisector  of  a  chord 
of  a  circle  will  pass  through  the  centre  and  will  bisect  the 
arcs  subtended  by  the  chord. 

Section  20  furnishes  the  proof  for  the  first  part  of  the 
theorem. 


64  ELEMENTS   OF   GEOMETRY. 

For  the  establishing  of  the  second  part,  the  analysis 
suggests  that  we  revolve  one  por- 
tion of  the  figure  on  PQ  as  an  axis. 
PQ  is  the  diameter  of  the  circle 
that  is  the  perpendicular  bisector 
of  AB. 

When  revolved,  MA  will  coincide 
with  MB,  and  the  point  A  will  fall 
at  B.     The  points   Q  and  P  will 

...  Fio.  64. 

remain  stationary. 

The  two  circumferences  will  coincide,  for  every  point 
of  each  will  be  at  the  same  distance  from  C 

Hence  QA  will  coincide  with  QB,  and  PA  will  coincide 

with  PB. 

Note.  —  The  chord  AB  subtends  the  two  arcs  AQB  and  APE. 
But  ordinarily,  in  speaking  of  the  arc  subtended  by  a  chord,  the 
lesser  arc  is  the  one  understood. 

Exercises.  —  1.  Show  that  chords  A  Q  and  BQ  would  be  equal 
to  each  other  ;  and  that  chords  AP  and  BP  would  also  be  equal  to 
each  other. 

2.  Show  how  to  draw  a  tangent  at  a  given  point  of  a  circumfer- 
ence. 

8.  Having  given  a  circumference,  show  how  to  find  the  centre. 

46.  Recalling  the  matter  in  §§  7, 11,  and  21,  and  again 
observing  the  generation  of  an  angle  by  the  rotation  of  a 
line  about  one  of  its  points  as  a  pivot,  we  are  prepared 
to  develop  another  relation ;  viz.  the 

Theorem.  If  a  circumference  be  constructed  with  the 
vertex  of  an  angle  as  its  centre,  the  arc  included  between  the 
lines  forming  the  angle  will  be  the  same  fractional  part  of 
the  entire  circumference  that  the  angle  is  of  360°. 


CIRCLES.  65 

During  a  complete  rotation  every  point  in  the  line  AB 
will  generate  a  circumference.  Let  B  represent  one  of 
these  points. 

If  the  angular  magnitude  about  A  be  generated  by  a 
uniform  rotation  of  the  line  AB,  any  point  in  the  line 
AB  will  move  at  a  uniform  rate. 


/ 

/ 

y^c 

/ 

A^ 

\ 

1      — 

w 

^- 

> 

^^    / 

/ 

/ 

/ 

Fig.  65. 

The  angle  and  the  circumference  are  generated  with 
uniformity ;  they  are  begun  at  -the  same  instant ;  and  the 
360°  of  rotation  are  completed  at  the  instant  the  circum- 
ference is  completed.  At  any  stage  of  the  proceeding, 
therefore,  the  angle  generated  will  be  the  same  fractional 
part  of  360°  that  the  arc  generated  by  any  point  is  of 
the  entire  circumference,  or 

6°  BG 


360°     circum. 

Corollary.  Tivo  angles  having  their  vertices  at  the 
centre  of  a  circle  will  have  the  same  ratio  as  the  intercepted 
arcs. 

e°  BC 


360°     circum/ 


(1) 


56  ELEMENTS   OP   GEOMETRY. 

Dividing  the  members  of  (1)  by  the  members  of  (2), 
we  get 

er  ^  BC 

*^°       CD 

Q.  E.  D. 

Notes.  —  1.  A  corollary  is  a  subsidiary  theorem  that  follows 
from  a  principal  one.  In  this  work  there  are  very  few  corollaries 
presented ;  it  being  preferred  to  establish  the  facts  and  relations 
as  original  exercises. 

2.  Because  of  the  fact  that  when  the  vertex  of  an  angle  is  at 
the  centre  of  a  circle,  its  sides  intercept  the  same  fractional  part 
of  the  circumference  that  the  angle  is  of  360°,  we  say  that  the 
angle  is  measured  by  the  intercepted  arc. 

For  purposes  of  numerical  description,  in  speaking  of  a  single 
angle  when  the  angle  itself  is  not  represented  in  the  drawing,  an 
angle  of  1°  is  applied  as  many  times  as  it  will  be  contained  in  the 
angle  ;  then  the  one  sixtieth  part  of  one  degree  (or  an  angle  of  one 
minute)  is  applied  to  the  remainder  as  many  times  as  it  wUl  be 
contained  in  it ;  then  to  the  second  remainder  is  applied  an  angle 
of  one  second ;  and  if  a  nearer  approximation  is  desired,  the 
decimal  subdivisions  of  a  second  are  applied  to  the  succeeding 
remainders  until  the  numerical  description  is  as  accurate  as  the 
circumstances  demand. 

In  the  same  way,  when  numerical  description  is  needed  in  order 
to  represent  an  arc  of  a  circumference,  an  arc  which  subtends  an 
angle  of  1°  is  applied  as  far  as  possible  ;  to  the  first  remainder  is 
applied  an  arc  that  subtends  an  angle  of  1',  until  there  is  a 
remainder  less  than  1' ;  to  this  is  applied  the  arc  that  subtends 
1",  etc. 

Ordinary  surveying  instruments  describe  an  angle  to  vdthin  30"; 
very  accurate  geodetic  instruments  to  within  10"  ;  ordinary  astro- 
nomical instruments  to  within  3  " ;   and  the  best  to  within  ^". 


CIRCLES. 


67 


47.  Definitions.  If  two  chords  intersect  on  the  circum- 
ference of  a  circle,  the  angle  they  make  is  said  to  be*  an 
inscribed  angle,  and  the  intercepted  arc  is  said  to  subtend 
the  angle. 

Theorem.  An  inscribed  atigle  is  measured  by  half  of 
the  intercepted  arc. 

c 

(a)  If  one  of  the  chords  be  a  diameter  as  AB,  draw  the 
auxiliary  line  CD. 


CA=CI). 
.'.ZCAD  =  ZCDA. 
But     Z  BCD  =  ZCAD+ZCDA 
=  2Z  GAD. 

The  Z  BCD  is  measured  by  the 
arc  BD.  Hence  the  Z  BAD  is 
measured  by  one-half  the  arc  BD. 

(b)  If  the  centre  be  within  the 
angle  formed  by  the  two  chords, 
draw  an  auxiliary  diameter  and 
then  demonstrate. 

(c)  If  the  centre  be  without  the 
angle  formed  by  the  two  chords, 
draw  an  auxiliary  diameter  and 
demonstrate. 


Fig.  6T. 


The  three  cases  are  all  the  possible  ones,  and  each 
having  been  demonstrated,  the  theorem  is  established. 


58 


ELEMENTS    OF   GEOMETRY. 


Ezercises. —  1.  Show  that  an  inscribed  right  angle  will  be  sub- 
tended by  a  semicircumference  ;  an  inscribed  acute  angle  by  an 
arc  less  than  a  semicircumference  ;  and  an  inscribed  obtuse  angle 
by  an  arc  greater  than  a  semicircumference. 

2.  A  secant  separates  a  circle  into  two  parts  called  segments. 
Show  that  all  angles  inscribed  in  a  given  segment  are  equal. 


Fig.  68.  ' 

3.  Show  that  angles  inscribed  in  the  two  segments  formed  by  a 
secant  will  be  supplementary. 


Pio.  69. 


4.  Show  that  the  angles  formed  by  a  tangent,  and  a  secant 
passing  through  the  point  of  tangency,  are  measured  by  the  halves 
of  the  intercepted  arcs. 


CIKCLES. 


59 


Fig.  70. 


48.  Theorem.  Parallel  secants  intercept  equal  arcs  of 
the  circumference  of  a  circle. 

If  AG  and  BD  are  equal,  they  may  be  brought  to 
coincidence  by  rotation  about  a 
diameter    perpendicular    to  the  i 

parallel  chords. 

Acting  upon  this  suggestion, 
draw  a  perpendicular  through  O. 
It  will  be  a  perpendicular  bisector 
of  both  chords. 

Revolve  one   semicircle  upon 
the  diameter  as  an  axis.     All  the 
parts  of  the  revolved  figure  will 
coincide  with  the  parts  of  the  figure  that  remains  sta- 
tionary.    Hence  AG  =  BD. 

Exercises.  —  1.  Demonstrate  the  theorem,  making  use  of  the 
principles  establislied  in  §  45. 

2.    Establish  it  by  the  principles  of  §  47. 

49.  Theorem.  An  angle  formed  by  two  secants  which 
intersect  within  a  circle,  is  measured  hy  the  half-sum  of  the 
arcs,  subtended  by  the  angle  considered,  and  its  vertical 
angle. 

Let  AC  and  BD  represent  any  two  secants  fulfilling 
the  required  conditions,  and  0 
the  angle  considered. 

Analysis.  — If  a  line  be  drawn 
parallel  to  BD,  it  will  make 
with  AC  an  angle  6 ;  and  if  it 
be  drawn  so  as  to  intersect  the 
circumference,  it  will  intercept 
equal  arcs. 


60 


ELEMENTS   OF   GEOMETRY. 


Demonstration.  —  Through  A  draw  a  parallel  to  BD. 
Z  CAE  =  6, 
Eb  =  AB, 
dE=cb  +  AB, 


e  =  ZCAE  =  f  =  2^^- 


Q.  E.  D. 


Ezercise.  —  Show  the  truth  of  the  theorem  by  drawing  auxiliary 
lines  through  the  centre  parallel  to  the  secants. 

60.   Let  AC  and  BD  intersect  within  the  circle.     The 
ZOis  measured  by  the  half-sum 
of  the  arcs  AB  and  CD. 

The  directions  indicated  by 
the  arrowheads  are  positive. 

If  the  secant  BD  be  moved 
parallel  to  its  initial  position, 
so  that  CD  be  increased,  AB 
will  be  diminished  by  a  like 
amount. 

When  the  position  AD'  shall 
have  been  reached,  the  Z  6  will 
be  measured  by  the  half  of  CD'. 

If  a  further  movement  be  made,  under  the  same  condi- 
tions, and  the  position  ED"  be  reached,  the  Z  6  will  not 
have  changed,  the  arc  measured  from  C  will  have  increased, 
but  the  arc  AG  will  be  measured  in  a  reverse  sense  from 
the  arc  AB,  and  is  negative.  In  length  it  is  equal  to  the 
arc  D'D",  but  imder  the  circumstances  is  negative. 

Hence  the  angle  CED"  is  measured  by  the  half-sum  of 
the  intercepted  arcs,  —  the  arc  that  is  convex  toward  the 
point  E,  being  negative,  and  the  one  concave  toward  E 
being  positive. 


Fig.  72. 


CIRCLES. 


61 


If  the  secant  be  further  moved  until  it  becomes  a 
tangent,  it  is  readily  seen,  by 
drawing  AK  parallel  to  the 
tangent  line,  that  the  Z  CFT 
is  measured  by  half  the  aggre- 
gate of  the  two  arcs  GT  and 
AT;  the  arc  AT  being  nega- 
tive. 

If  the  secant  FC  should  now 
move  parallel  to  itself  until 
it  shoidd  become  a  tangent, 
the  angle  which  has  remained 
the  same  would  be  measured 
by  the  aggregate  of  the  arcs 
RMT  and  RQT,    the  latter  being  negative. 

The  student  should  show  this  by 
draAving  an  auxiliary  line  through 
one  point  of  tangency  parallel  to  the 
other  tangent,  or  in  any  other  way 
that  he  may  choose. 


Fig.  73. 


Fig.  74. 


51.  Theorem.  In  the  same  or  in 
equal  circles,  equal  chords  subtend  equal 
arcs. 

Analysis.  —  If  the  equal  chords  do 
subtend  equal  arcs,  the  angles  formed 
by  joining  their  extremities  with  the  centre  will  be  equal, 
because  angles  at  the  centre  are  measured  by  the  inter- 
cepted arcs. 

Proof.  —  Draw  the  auxiliary  lines  indicated. 
The  A  ABC  and  DEC  are  equal,  having  the  three 
sides  of  one  equal  to  the  three  sides  of  the  other. 


62 


ELEMENTS   OF   GEOMETRY. 


..ZACB  =  ZDCE, 

and  becaiise  the  angles  are 
equal,  the  arcs  subtending 
them  Avill  be  equal. 

Q.  E.  D. 

Exercise.  —  1.  It  is  said  that 
equal  chords  are  equally  distant 
from  the  centre.     Is  it  true  ?  Fio.  75. 

2.    Demonstrate  the  converse  of  the  theorem. 

8.    Demonstrate  the  opposite  of  the  theorem. 

52.  Theorem.  In  a  circle  the  greater  of  two  chords 
subtends  the  greater  arc. 

Analysis. — Jj^the  greater  chord  does  subtend  the  greater 
arc,  when  the  arcs  are  brought        --.^ 
BO  as  to  have  two  extremities 
coincident,  the  point  D  will  lie 
beyond  B  from  A. 

Demonstration. — The  point  D 
does  lie  beyond  B;  for  if  we 
construct  an  auxiliary  circle 
having  A  for  its  centre  and 
AB  for  its  radius,  the  point  D 
will  fall  outside  of  the  auxiliary 
circle,  for  by  hypothesis  the 
distance  AD  is  greater  than  the  radius  of  the  auxiliary 
circle. 

Then  the  arc  which  is  subtended  by  the  greater  chord 
will  lie  upon  the  arc  subtended  by  the  lesser  chord 
throughout  its  entire  length  and  extend  beyond  it. 
Hence  it  is  greater,  and  the  theorem  is  established. 


Fig.  76. 


CIRCLES.  63 

53.  Theorem.  Tlie  lesser  of  tivo  unequal  chords  in  a 
circle  will  be  at  the  greater  distance  from  the  centre. 

The  distances  of  the  chords  from  the  centre  are  the 
lengths  of  the  perpendiculars  from 
the  centre  to  the  chords. 

If  the  chords  be  made  to  have 
an  extremity  of  each  coincident, 
the  chords  themselves  will  not 
coincide,  but  will  occupy  the  relar 
tive  positions  indicated  in  the  fig- 
ure. AD,  being  the  greater  chord, 
siibtends  the  greater  arc,  and  so 
lies  in  such  a  position  that  any  line  drawn  from  C  to  any 
point  in  AB  will  cross  AD.  Therefore  the  ±  CH  will 
cross  AD  at  some  point,  as  P,  which  does  not  coincide 
with  K. 

.\CH>CP>CK 

(read:    CH  is  greater  than  CP,  which  is  greater  than 

CK).  Q.  E.  D. 

Exercises.  —  1.  All  chords  of  equal  length  in  a  given  circle  are 
equally  distant  from  the  centre. 

2.  Find  the  maximum  and  minimum  chords  that  may  be  drawn 
through  a  given  point  in  a  circle. 

3.  Establish  the  converse  of  the  theorem. 

4.  Find  the  locus  of  any  fixed  point  on  a  chord  of  given  length. 

54.  Theorem.  If  two  circumferences  intersect,  the  line 
of  centres  ivill  be  the  perpendicular  bisector  of  their  common 
chord. 

Analysis. — If  the  line  00'  be  the  perpendicular  bisector 
of  the  chord  CH,  it  must  contain  at  least  two  points 
which  are  equally  distant  from  C  and  H.     Does  it  ? 


64 


ELEMENTS    OP    GEOMETRY. 


Demonstration. — 0  is  equally  distant  from  C and  ^;  and 
()'  is  equally  distant  from  C  and  H.     Hence  the  theorem. 


Fig.  78. 

Ezerclses.  —  1.  Show  that  if  two  circles  intersect,  the  distance 
between  their  centres  is  less  than  the  sum  of  their  radii. 

2.  Show  that  if  two  circles  are 
tangent  to  each  other,  they  wUl 
have  a  common  tangent  line. 

8.  Show  that  the  line  of  centres 
will  pass  through  the  point  of  tan- 
gency,  and  will  equal  the  sum  of 
the  radii.  

4.  Show  that  if  two  circles  are  ^'®-  ^®- 

exterior  the  one  to  the  other,  the  distance  between  centres  will  be 
greater  than  the  sum  of  their  radii. 

65.   Definitions.    A  triangle  is  inscribed  within  a  circle 
when  its  vertices  are  on  the  circumference  and  its  sides 


Fio.  80. 


CIRCLES.  65 

are  chords.     The  circle  is  said  to  be  circumscribed  about 
the  triangle. 

A  circle  is  inscribed  within  a  triangle  when  the  sides  of 
the  triangle  are  tangent  to  the 
circumference  and  the  circle  lies 
within  the  triangle. 

A  circle  is  escribed  to  a  tri- 
angle when  the  lines  forming 
the  triangle  are  tangent  to  the 
circumference,  but  the  circle 
does  not  lie  within  the  triangle. 

An  inscribed  polygon  is  a  poly- 
gon the  vertices  of  which  lie  on 
the  circumference,  and  the  sides  of  which  are  chords. 

GENERAL   EXERCISES. 

1.    Problem.     Through  a  point  without  a  circle  to  draw  a 
tangent  to  the  circumference. 


Analysis.  —  If  PT  were  the  required  tangent  through  P,  it 
would  be  perpendicular  to  a  radius  drawn  to  the  point  of  tangency. 
If  then  we  had  CT  drawn,  Z  CTP  would  be  90°.  If  we  had  a 
line  CP,  joining  the  two  fixed  points,  a  right  triangle  would  be 


66  ELEMENTS   OF    GEOMETRY. 

formed  having  CP  for  its  hypothenuse  ;  and  if  a  circle  were  con- 
structed with  CP  as  its  diameter,  the  circumference  would  pass 
through  T. 

Bemark. — By  the  analysis  a  sufficient  number  of  necessary 
and  previously  established  relations  have  been  determined  to  enable 
us  now  to  make  the  construction  which  in  its  order  will  be  the 
reverse  of  the  analysis. 

Construction.  — Join  the  centre  of  the  given  circle  to  the  given 
I)oint.    On  this  segment  as  a  diameter  construct  the  circumfer- 


Fi6.  8S. 

ence  of  a  circle.  It  intersects  the  given  circumference  at  T  and  T. 
Draw  PT  and  PT'.  Both  will  be  tangents,  because  they  will  be 
perpendicular  to  radii  drawn  from  C  to  T  and  T'.  Z  PTC  and 
ZPVC  are  each  inscribed  in  semicircumferences.  Hence  we 
have  two  constructions. 

Discussion.  —  It  is  to  be  observed  that  if  the  point  P  should 
move  to  a  greater  distance  from  the  given  circle,  the  tangents 
would  approach  parallelism. 

If  the  point  P  should  move  to  a  position  on  the  circumference, 
the  two  tangents  would  coincide  and  form  one. 

If  the  point  P  were  within  the  circle,  the  construction  would  not 
be  possible. 


CIRCLES. 


67 


2.  Show  that  of  all  the  points  on  the  circumference  of  a  circle, 
the  nearest  and  the  furthest  from  any  given  point  will  be  on  the 
line  joining  the  given  point  with  the  centre  of  the  given  circle. 


Fig.  84. 


8.  Show  that  of  all  the  points  on  the  circumference  of  a  circle, 
the  nearest  and  the  furthest  from  another  circumference  will  be  on 
the  line  joming  theu'  centres. 


Fig.  85. 


4.  Construct  a  circle  of  given  radius  that  shall  be  tangent  to  two 
given  circles. 

Note.  —  Make  as  many  constructions  as  possible,  and  discuss 
the  limitations  of  each. 

6.  Show  that  if  from  a  point  without  a  circumference  tangents 
be  drawn,  the  line  joining  the  point  with  the  centre  bisects  the 
angle  formed  by  the  tangents;  bisects  the  angle  formed  by  the 
radii  drawn  to  the  points  of  tangency ;  and  bisects  the  arcs.  The 
segments  of  the  tangents  are  equal. 

7.  Inscribe  a  circle  within  a  given  triangle. 

8.  Use  a  triangle  inscribed  within  a  given  circle  to  show  that  in 
any  triangle  having  unequal  sides  the  greater  of  two  sides  will  lie 
opposite  the  greater  angle. 


68 


ELEMENTS   OF   GEOMETRY. 


9.  Show  that  in  a  right  triangle  the  difference  between  the  sum 
of  the  perpendicular  sides  and  the  hypothenuse  will  be  the  diam- 
eter of  the  inscribed  circle. 


Fie.  8«. 

10.  Show  that  the  bisector  of  an  interior  angle  and  the  bisectors 
of  the  non-adjacent  exterior  angles  will  pass  through  one  point, 
and  that  point  will  be  the  centre  of  an  escribed  circle. 

11.  Show  that  if  a  regular  hexagon  be  inscribed  within  a  circle, 
each  side  will  equal  the  radius  of  the  circle. 

12.  Show  that  if  any  quadrangle  be  inscribed  within  a  circle, 
the  opposite  angles  will  be  supplementary. 

13.  Prove  the  converse. 

14.  Construct  a  segment  of  a  given  circle  that  shall  be  capable 
of  containing  a  given  angle. 


Fio.  87. 


16.  With  a  given  line  as  a  chord  construct  a  segment  of  a 
circle  that  shall  contain  a  given  angle. 


CHAPTER   V. 

56.  Definitions.  If  a  point  move  along  a  line,  as  AB, 
from  any  point,  as  A,  toward  B,  it  will  in  its  course 
occupy  the  position  of  every  point  of  the  line  as  far  as 
we  may  conceive  it  as  moving. 


3  3  B  i 

■H 1 h-»- 

FiG.  88. 


If  we  fix  our  attention  upon  B,  and  speak  of  the  dis- 
tance of  B  from  A,  it  is  a  definite  thing,  and  is  perfectly 
understood. 

For  purposes  of  description  and  comparison  we  fre- 
quently take  some  convenient  unit  of  measure,  and  apply 
it  to  the  distance.  If  the  distance  be  a  day's  journey,  we 
use  the  mile  or  the  kilometre.  If  it  be  a  distance,  as  in 
the  figure,  between  points  on  the  page  of  this  book,  we 
use  inches  or  centimetres. 

Eemembering  that  the  point  in  moving  along  the  line 
from  A  to  B  occupies  an  infinite  number  of  positions, 
one  sees  that  the  chances  that  the  extremity  of  the 
measuring  unit  will  not  fall  at  B  are  as  infinity  ( oo  )  to 
one. 

69 


70  ELEMENTS   OF   GEOMETRY. 

In  the  above  figure,  we  would,  "  roughly  speaking,"  say- 
that  B  was  3  centimetres  from  A.  If  we  desired,  for  any 
reason,  more  accurately  to  describe  the  distance,  we  should 
descend  to  fractional  parts  of  this  unit;  the  fractional 
part  being  less  than  the  distance  by  which  we  failed  to 
reach  B  when  using  the  entire  unit.  Again,  the  chances 
are,  as  infinity  to  one,  that  the  new  unit  of  measure  will 
not  fall  on  B.  The  subdividing  of  the  unit  may  be  car- 
ried to  any  extent,  depending  entirely  upon  the  required 
accuracy  of  description. 

The  most  convenient  subdivision  is  the  decimal  one. 

If  the  applied  unit  or  any  of  its  subdivisions  have 
their  extremities  at  B,  the  distance  AB  and  the  unit  are 
said  to  be  commensurable.  In  general,  however,  if  a  point, 
as  B,  is  taken  at  random  on  the  line  AB,  its  distance  from 
A  will  not  be  commensurable  in  terms  of  any  established 
unit.  The  distance  is  then  said  to  be  incommensurable 
with  the  unit. 

It  might  be  commensurable  with  respect  to  one  unit 
and  incommensurable  with  respect  to  another  unit.* 

A  point  may  be  so  assumed  that  its  distance  from  A 
shall  be  commensurable  in  terms  of  any  unit  that  may  be 
selected. 

♦  Note  in  Illustration.  —  If  we  undertake  to  express  deci- 
mally the  distance  of  a  point  from  A  that  is  distant  therefrom  |  of 
the  unit  distance,  we  can  only  approximate  to  it.  The  first  approx- 
imation would  be  .6,  a  nearer  one  would  be  .66,  a  still  nearer  .666. 
"We  might  continue  annexing  decimal  places  as  long  as  we  please 
and  we  should  never,  in  that  way,  reach  the  point  that  is  |  of  a 
imit's  distance  from  A ;  although  at  each  step  we  should  come 
nearer  the  point. 

I  is  said  to  be  the  limit  toward  which  we  approacli  as  we 
increase  the  number  of  decimal  places  in  .6666 


A 

B 

h 

C 

D 

AREA.  71 

57.  We  know  from  §  19  that  parallel  lines  are  every- 
wliere  equally  distant  from  each  other.  Let  AB  and  CI) 
in  the  figiu-e  represent  two  parallel  lines,  and  AC  a,  line 
perpendicular  to  AB  and  CD.  Represent  the  segment 
^Cby(^). 

If  we  cause  the  line  AC  to 
move  parallel  to  itself  a  dis- 
tance (6),  the  extremities  of  (K) 
remaining  in  AB  and  CD,  the 
surface  swept  over  by  the  seg- 
ment (h)  is  described  as  'Hhe  ^^^'  ^* 
area  bh." 

AC  and  BD  are  parallel,  and  if  the  line  CD,  perpen- 
dicular to  them  and  remaining  always  parallel  to  its 
initial  position,  should  move  to  the  position  AB,  the 
segment  (b)  would  sweep  over  the  area  (hb).  The  areas 
swept  over  being  the  same,  we  have  bh  =  hb. 

Either  side  of  the  rectangle  ACDB  may  be  called  the 
base ;  ordinarily  it  would  be  the  lower  one  in  the  figure. 
The  perpendicular  distance  between  the  base  and  its 
parallel  is  called  the  altitude.  Sometimes  this  parallel 
is  called  the  upper  base. 

The  rectangle  is  accurately  described  by  bh  or  7ib. 

In  general,  b  and  h  are  incommensurable  with  any 
assumed  unit  of  length,  and  the  area  bh  is  incommensu- 
rable with  the  square,  having  that  unit  for  its  side.  But 
for  convenience  of  nvimerical  description  or  comparison 
some  unit  square  is  taken  and  applied  to  the  rectangle. 
If  it  be  large  tracts  of  land  that  are  being  considered,  we 
use  the  acre  or  the  hectare.  If  it  be  the  areas  of  rect- 
angles on  a  page  of  this  book,  the  square  inch  would  be 
appropriate. 


72 


ELEMENTS   OF   GEOMETRY. 


A 

B 

H 

N 

h 
M 

C 

1 

i 

K 

D 

Fig.  90. 


If  we  should  assume  CM  as  the  side  of  a  unit  square, 
and  should  lay  it  off  from  C  towards  A  as  many  times  as 
possible,  an  extremity  would  fall  within  a  unit's  length 
of  Af  as  at  H.  If  we  lay  off  the  same  unit  of  length 
from  C  toward  D,  as  many  times  as  possible,  an  extrem- 
ity would  fall  within  a 
unit's  length  of  D,  as  at  K. 

If  the  line  C-£f  move  par- 
allel to  itself  and  so  that 
each  point  moves  on  a  per- 
pendicular to  CA,  until  it 
occupy  the  position  KN, 
it  mil  have  swept  over  an 
area  expressed  by  CK  x  KN.  It  will  be  commensurable 
with  the  assumed  unit  area,  and  the  number  of  times  it 
will  contain  that  unit  is  expressed  by  the  product  of  the 
number  of  units  of  length  in  CH  by  the  number  of  units 
of  length  in  CK. 

If  a  nearer  approximation  is  desired,  the  former  unit 
of  area  is  subdivided,  so  that  a  side  of  the  new  compari- 
son square  will  be  less  than  the  distance  by  which  in  the 
preceding  instance  we  failed  of  reaching  either  AB 
or  BD. 

We  shall  thus  have  an  increased  commensurable  area, 
which  will  be  a  nearer  approximation  to  the  area  hh. 

This  subdivision  of  the  unit  after  the  manner  above 
indicated  may  be  carried  as  far  as  we  please,  and  a  com- 
mensurable area  be  expressed  in  the  terms  of  some 
measiiring  unit,  which  shall  approximate  as  nearly  as  we 
may  please  to  the  incommensurable  area  bh. 

hh  is  not  necessarily  incommensurable,  but  is  gen- 
erally so. 


AREA. 


73 


58.  Let  A  i  3present  the  area  of  the  rectangle  CDEF, 
and  ^1'  the  area  of  the  rectangle  HJKL.  Then  A  =  hh, 
and  A'  =  h'h'. 

Dividing  the  members  of  one  equation  by  the  members 
of  the  other,  we  have,     A  _  hh 

A'^b%'' 
i.e.  two  rectangles  have  the  same  ratio  as  the  products  of 
their  bases  and  altitudes. 


F 

E 

h 

b 

C 

D 

L 

K 

iC 

b' 

Hi 

J 

Fig.  91. 

If  h  and  h'  happen  to  be  equal, 
A_b 
A'~b'' 
i.e.  when  the  altitudes  are  equal,  the  ratio  of  the  areas  is 
the  same  as  the  ratio  of  the  bases. 

If  instead  of  h  and  h'  being  equal,  it  happen  that  b 
and  b'  were  equal,  we  would  have, 

A'~h'' 
i.e.  the  ratio  of  the  areas,  when  the  bases  are  equal,  would 
be  the  same  as  the  ratio  of  the  altitudes. 

59.  Let  ABCD  represent  any  parallelogram.  If 
through  a  pair  of  adjacent  vertices,  as  A  and  D,  lines 
be  drawn  perpendicular  to  the  line  AD,  a  rectangle 
AKHD  will  be  formed.  The  area  ABHD  is  common 
to  the  rectangle  AKHD  and  the  parallelogram  ABCD. 
The  parts  of  each  not  common  to  the  other  are  the  tri- 


74 


ELEMENTS   OF   GEOMETRY. 


angles  AKB  and  DHC.     These  triangles  ire  congruent, 
therefore  the  surface  AKHD  equals  the  surface  ABCD. 


Fig.  92. 

The  surface  AKHD  =  bh.  Therefore  the  surface 
ABCD  =  bh,  in  which  b  is  one  side  of  the  parallelogram, 
and  h  is  the  segment  of  the  perpendicular  to  (b)  that  is 
included  between  (6)  and  the  line  forming  the  opposite 
side  of  the  parallelogram.      Therefore, 

The  surface  of  a  parallelogram  will  be  represented  by  bh. 

Exercises.  —  1.  Show  that  the  ratio  of  the  surfaces  of  two 
parallelograms  to  each  other  equals  the  ratio  of  the  products  of 
bases  and  altitudes. 

2.  If  the  altitudes  are  equal,  the  surfaces  will  have  the  same 
ratio  as  the  bases. 

3.  Show  that  if  lines  be  drawn  through  two  vertices  of  a  tri- 
angle parallel  to  the  opposite  sides,  a  parallelogram  will  be  formed, 
the  surface  of  which  will  be  double  that  of  the  triangle. 

60.  By  the  last  exercise  in  the  preceding  article  it  is 
shown  that  a  triangle  will  be  half  of 
a  certain  parallelogram.  By  §  59,  any 
parallelogram  is  equivalent  to  a  rect- 
angle having  the  same  base  and  alti- 
tude, and  the  area  of  a  rectangle  is 
represented  by  (bh),  in  which  (b)  represents  the  base,  and 
(Ii)  represents  the  altitude.     Therefore, 


The  surface  of  a  triangle  will  be  represented  by 


bh 


PKOPORTIONAL  DIVISION.  75 

Exercises.  —  1.  Show  that  the  ratio  of  the  surfaces  of  any  two 
triangles  to  eacli  other  equals  the  ratio  of  the  products  of  their 
bases  and  altitudes. 

2.  Show  that  if  their  bases  are  equal,  the  surfaces  will  be  to 
each  other  as  their  altitudes. 

3.  Show  that  if  their  altitudes  are  equal,  the  surfaces  will  be  to 
each  other  as  their  bases. 

4.  Show  that  if  two  triangles  have  their  bases  in  the  same  line 
and  their  vertices  at  the  same  point,  their  areas  are  to  each  other 
as  their  bases. 

5.  Show  that  if  two  triangles  have  their  bases  in  the  same  line 
and  their  vertices  in  a  parallel  line,  the  ratio  of  their  areas  equals 
the  ratio  of  their  bases. 

61.   Theorem.     The  area  of  a  trapezoid  equals  the  half- 
sum  of  its  parallel  sides  multi- 
plied by  the  perpendicular. 


Let    ABCD    represent    the 
trapezoid.     Draw  a  diagonal  as      7-^  -D^ 

AC.     Each  triangle  composing  ^lo-  ^^• 

the  trapezoid  will  have  the  same  altitude  (Ji)  ;  hence, 

AxQ?^  ABC  =  ^  BC '  h. 
Avesi  ACD  =  ^  AD  •  7i. 
ABC  +  ACD=:^  (BC)  'h-\-^  (AD)  •  h. 
But  ABC  +  ACD  =  area  of  the  trapezoid. 

.'.  Avesi  ABCD  =  ^  (BC  +  AD)  .  h.  •       q.e.d. 

Exercise.  —  Show  that  the  segment  of  the 
line  joining  the  middle  points  of  the  non-parallel 
sides  of  a  trapezoid  will  equal  the  half-sum  of 
the  parallel  sides.  ^i®*  ^^- 


76  ELEMENTS   OF   GEOMETRY. 

62.  Theorem.  If  a  line  be  draivn  parallel  to  any  side 
of  a  triangle,  the  other  sides  will  be  separated  into  segments, 
which  will  form  equal  ratios. 


^7\  £\ 

Fig.  96. 

Let  ADE  represent  the  given  triangle,  and  BC  a  line 
drawn  parallel  to  DE.  Draw  the  auxiliary  lines  through 
the  vertices  parallel  to  the  opposite  sides. 

By  Exercise  2,  §  59, 

UABKH     AB 


(1) 
(2) 


EJADEH     AB 

OACQG^AO 

OAEDG     AE 

But  O  ABKH  ^CJ  ACQB,  since  their  bases  BK  and 

QC  are  each  equal  to  DE,  and  their  altitudes  are  the 

same.     Also  CJ  ADEU =EJ  AEDG,  since  they  have  the 

same  base  and  equal  altitudes. 

The  first  members  of  equations  (1)  and  (2)  being  equal, 
the  second  members  are,  and 

AB     AC  ^ „ ^ 

AD     AE 

Exercises.  —  1.  Show  that  — =  — 
BD      CE 

2.   Show  that  ^  =  :d^. 
AC     AE 

8.   Show  that  ^  =  ^. 
AC     CE 


PROPOKTIONAL   DIVISION. 


77 


4.  Show  that  if  a  line  be  drawn  through  the  middle  point  of  one 
side  of  a  triangle,  parallel  to  a  second  side, 
it  will  bisect  the  third  side. 

5.  Show  that  the  new  triangle  formed  in 
Ex.  4  will  be  one-fourth  the  area  of  the 
original  triangle.  P^^  g- 

63.  Problem.  Establish  the  converse  of  the  theorem 
in  §  62,  viz. :  If  a  straight  line  be  drawn  through  points 
that  separate  two  sides  of  a  triangle  into  proportional  * 
segments,  it  will  be  parallel  to  the  third  side. 

Recalling  the  fact  that  the  reductio  ad  absnrdum  is  a 
particularly  appropriate  method  for  de- 
termining the  truth   or   falsity   of  the 
converse  of  a  theorem,  we  proceed  as 
follows : 

If  the  line  BC  be  the  line  through 
the  points  of  proportional  division,  and 
if  it  be   not  parallel   to  DE,  let   us   draw   a  line  BQ 
through  B  that  shall  be  parallel  to  DE.     By  §  62, 
AB^AQ 
AD     AE 

AB     AC 

But = by  hypothesis.    Multiplying  each  mem- 
ber of  both  equations  by  AE,  we  have,  from  (1), 
AB 
AD 
AB 


(1) 


AE  =  AQ, 


from  (2), 


AD 


-AE  =  Aa 


*  Note.  —  Four  quantities  which  may  form  two  equal  ratios  are 
proportional.  Frequently  the  numerators  of  the  fractions  used  in 
expressing  a  proportion  are  called  antecedents,  and  the  denomina- 
tors consequents. 


78  ELEMENTS   OF   GEOMETRY. 

The  first  members  are  the  same,  hence  the  second 
must  be,  and 

AQ  =  AC. 

The  supposition  that  the  line  BC  is  not  parallel  to  DE 
thus  leads  us  to  an  absurdity,  and  the  supposition  that 
BO  is  not  parallel  to  DE  is  an  erroneous  one. 

64.  Theorem.  If  two  triangles  have  the  three  angles  of 
one  equal  to  the  three  angles  of  the  other,  each  to  each,  the 
ratio  of  any  two  sides  of  one  will  equal  the  ratio  of  the 
cori'esponding  *  sides  of  the  other. 


Fig.  99. 

Let  ABQ  and  DEF  represent  two  mutually  equi- 
angular triangles. 

Let  ZABC=ZEDF, 

ZBAC=ZDEF, 

ZBCA^ZDFE. 

Superimpose  the  A  ABC  upon  the  A  EDF,  so  that  the 
Z  ABC  shall  coincide  with  the  Z  EDF.  The  interior 
angles  at  A  and  at  E  are  equal  by  hypothesis.  Then 
AC  is  parallel  to  EF. 

■■■M-w         ««^>'« 

♦Note.  —  In  mutually  equiangular  triangles,  the  sides  opposite 
equal  angles  are  called  "corresponding." 


PROPORTIONAL   DIVISION.  79 

If  the  triangles  had  been  superimposed  so  that  the 
interior  angles  at  C  and  F  had  been  made  coincident,  AB 
would  have  been  parallel  to  ED. 

■■■%=wi  (^  «')'('> 

Either  by  superposition  or  by  the  equality  axiom, 
using  equations  (1)  and  (2),  we  may  show  that 


BA^CA 
DE     FE 


Q.  E.  D. 


65.  Definitions.  Similar  figures  are  those  in  which  the 
angles  of  one  are  equal  to  the  angles  of  the  other,  and 
the  corresponding  sides  are  proportional. 

In  a  figure  other  than  a  triangle,  corresponding  sides 
are  best  described  as  being  those  that  lie  between  mutu- 
ally equal  angles. 

Exercises.  —  1.  Show  that  if  two  triangles  have  the  three  sides 
of  one  perpendicular  respectively  to  the  three  sides  of  another,  the 
two  triangles  will  be  mutually  equiangular,  and  hence  similar. 

2.  Show  that  if  two  triangles  have  an  angle  in  each  equal,  and 
an  angle  in  one  the  supplement  of  an  angle  in  the  other,  the  ratio 
of  the  sides  opposite  the  equal  angles  is  equal  to  the  ratio  of  the 
sides  opposite  the  supplementary  angles. 


Fig.  100. 


3.  Show  that  if  two  triangles  have  the  three  sides  of  one  parallel 
to  the  three  sides  of  the  other,  the  triangles  will  be  equiangular 
and  therefore  similar, 


80 


ELEMENTS   OF   GEOMETRY. 


66.   Theorem.     If  two  triangles  have  an  angle  in  each 
equal  and  the  including  sides  proportional,  they  are  similar. 


Let  the  interior  angles  at  A  and  D  be  equal,  and 

A.B     A.C 

——  =  — —  •     If  ABC  be  superimposed  upon  DEF  so  that 

DE     DF 

the  interior  angle  at  A  shall  coincide  with  the  interior 
angle  at  D,  we  shall  have  B  and  C  fall  at  points  on  DE 
and  DF  so  as  to  divide  them  proportionally. 
Then  by  §  63,  BC  will  be  parallel  to  EF,  and 

BC  ^AC 

EF        DF'      Q.E.D. 


Z  ABC  =  Z  DEF  ^^^  AB 
ZACB  =  ZDFE   ^    DE 


67.  Theorem.  If  the  three  sides  of  a  triangle  are  pro- 
portional to  the  three  sides  of  another  triangle,  they  will  he 
mutually  equiangular,  and  hence  similar. 


Fia.  102. 

This  is  the  converse  of  §  64. 

AB     AG      BC 


By  hypothesis, 


DE     DF     EF 


PKOPOKTIONAL   DIVISION.  81 

If  the  angles  are  not  equal,  when  we  attempt  super- 
position, causing  the  side  AB  to  fall  on  the  line  DE,  as 
at  DH,  the  vertex  C  will  fall  at  some  point  as  P,  not 
onZ)F. 

Through  the  point  fi,  draw  HK  parallel  to  EF,  and 
writing  DH  for  its  equal  AB,  we  have : 

By  §  62,  = 

•^         '  DE       DF 

But  by  hypothesis,  ^  =  ~ 


Also  by  §  62, 

But  by  hypothesis. 


.DK=AC=DP. 
DH     HK 


DE       EF 
DH     BC 


DE      EF 
..HK=BC=HP. 

Hence  the  A  DHP  and  DHK  having  three  sides  of 
one  equal  to  the  three  sides  of  the  other  are  equal  in  all 
their  parts,  and  Z  HDP  =  Z  HDK.  Therefore  the  sup- 
position that  the  vertex  C  did  not  fall  on  DF  is  an 
erroneous  one.  It  does  fall  on  DF,  and  §  66  applies  to 
establish  the  similarity.  q.  e.  d. 

68.  Constructions.  1.  To  divide  a  given  segment  of  a 
straight  line  into  a  given  number  of  equal  parts. 

Let  AB  represent  the  segment  that  is  to  be  separated 
into  equal  parts,  say  four. 

Draw  any  line  AK  in  a  convenient  position.  With 
the  dividers,  or  any  convenient  measure,  lay  off  a  con- 


ELEMENTS   OF   GEOMETRY. 


venient    distance  AJ.      Lay  off  JT=TL=LK=AJ. 
Join  KB.     Through  L,  T,  and  J  draw  parallels  to  KB. 


Fig.  103. 


By  §  62,  AB  will  be  separated  into  four  equal  parts. 

Q.  E.  p. 

2.    To  divide  a  given  segment  of  a  straight  line  into  parts 

that  shall  be  proportional  to  any  given  segments. 


Fig.  104. 

3.  To  draw  a  triangle  thai  shall  have  a  given  perimeter 
and  shall  be  similar  to  a  given  triangle. 

4.  To  find  a  fourth  proportional  to  three  given  segments 
of  straight  lines. 

If  X  represent  the  line  to  be  determined, 
a_c 
h~  X 


Fie.  IDS. 


MEDIANS. 


83 


The    form    immediately   suggests    an    application    of 
Exercise  2,  §  62. 


5.    To  construct  x  in  x  =  —• 
a 


X  b  ah 
I  =  ~>  or  -  =  -. 
0     a       0      X 


69.  Definition.  If  a  secant  of  a  triangle  pass  through 
a  vertex  and  the  middle  of  the  sides  opposite,  the  segment 
between  these  points  is  called  a  median,  There  will  be 
three  medians  in  every  triangle. 

Problems.  —  1.  Show  that  two  medians  of  a  triangle  trisect 
each  other;  i.e.  separate  each  other  into  two  segments,  one  of 
which  is  one-third  the  whole  median. 

The  analysis  of  the  problem  suggests 
the  following : 

Draw  PQ,  joining  the  middle  points 
of  AI  and  BI.     It  will  be  parallel  to 

TiA 

BA  and  equal  to  ^^. 
2 


Draw  DE;   it 
BA 


will  be  parallel  to  BA  and  equal  to  — — • 

Therefore  PQ  and  DE  are  parallel  and  equal,  and  PQED  is  a 
parallelogram  (Ex.  2,  §  38). 

Its  diagonals  bisect  each  other,  or  ^7=  ID.  But  QI=  AQhj 
construction ;  therefore  AQ  —  QI=  ID. 

For  the  same  reasons  BP  —  PI  —  IE.  q.  e.  d. 


84 


ELEMENTS   OF   GEOMETEY. 


8.   Show  that  the  third  median  would  also  pass  through  /. 

8.   Having  given  the  three  medians  of  a  triangle,  to  construct  it. 

Analysis. — If  ABC  were  the 
required  triangle,  with  m',  m", 
and  m'"  the  given  medians  ;  and 
if  we  should  draw  QE,  it  would 
be  parallel  to  BC  and  equal  to 


BG 

— .     If  BH  be  taken  equal  to 
2 

QB,  and  the  point  H  be  joined 
with  A,  P,  and  C,  three  parallelo- 
grams will  be  formed,  from  which 
we  may  establish  the  fact  that 
APH  will  be  a  triangle,  having  the  three  medians  for  its  sides. 

£'will  be  the  middle  point  of  AP. 

Construction.  —  Form  a  triangle  with  the   three    medians  as 
sides.    Draw  a  median  of  this  triangle  through  any  vertex.    Pro- 


Fi6.  109. 


Fi(i.  110. 


dues  it  one-third  of  its  length.  Through  either  of  the  other 
vertices,  as  A,  draw  AQ  and  AB,  and  lay  off  QB  =  AQ  and 
BC=AB.  Join^C.  ^PC  will  be  the 
required  triangle. 

4.  If  through  any  point  three  lines  be 
drawn  intersecting  parallels,  the  seg- 
ments will  be  proportional. 

5.  Establish  the  converse  of  Problem  4. 

6.  How  does  the  bisector  of  an  interior  angle  of  a  triangle 
divide  the  opposite  side  ? 


MEDIANS. 


85 


If  AD  bisects  the  interior  angle  at  A,  we  will  have  two  tri- 
angles, BAD  and  CAD  having  an  angle  in  each  equal,  and  other 
two  angles  supplementary. 


■^--^  K 


Fig.  112. 

By  Ex.  2.  §  65,  we  have  ^  =  ^. 
^  *  DC     AC 

Hence  the  opposite  side  is  divided  into  segments  proportional 
to  the  adjacent  sides. 

7.  How  does  the  bisector  of  an  exterior  angle  of  a  triangle 
divide  the  opposite  side  ? 

If  AK  bisects  the  exterior  angle  at  A,  we  have  two  triangles, 
BAR  and  CAR  having  the  ZK  in  common  and  the  Z  CAK 
supplementary  to  the  Z  BAK. 
BK^AB 
CK     AC 

Hence  the  two  segments  formed  by  the  point  of  intersection 
and  the  other  vertices  will  be  proportional  to  the  sides  having  their 
vertex  at  the  vertex  of  the  bisected  angle. 


By  Ex.  2,  §  65, 


8. 


Show  that  ^  =  :^. 
DC      CK 


GENERAL   EXERCISES. 

1.  On  a  given  segment  as  one  side  construct  a  parallelogram 
similar  to  a  given  parallelogram. 

2.  Show  that  similar  polygons  may  be  separated  by  auxiliary 
lines  into  similar  triangles. 

3.  Show  that  circles  are  similar 
figures. 

4.  Show  that  the  corresponding  alti- 
tudes of  similar  triangles  will  be  propor- 
tional to  any  set  of  cori-esponding  sides. 


Fig.  113. 


86  ELEMENTS   OF   GEOIVIETRY. 

5.  Show  that  the  radii  of  circles  inscribed  in  similar  triangles 
are  proportional  to  the  corresponduig  sides. 

6.  Show  that  the  same  relation  exists  between  the  diameters  of 
circumscribed  circles. 

7.  Show  that  the  same  relation  exists  between  the  radii  of  the 
corresponding  escribed  circles. 

8.  Show  that  the  corresponding  altitudes  are  to  each  other  as 
the  corresponding  medians. 

9.  Show  that  the  corresponding  angle  bisectors  are  to  each 
other  as  the  perimeters  of  the  triangles. 

Note.  —  There  are  three  principles  in  the  elements  of  geometry 
that  are  more  prominent  than  any  others. 

We  have  now  established  the  first  of  these,  as  follows : 

Corresponding  lines  of  similar  figures  are  to  each  other  as  ant 
OTHER  corresponding  lines. 

The  second  of  these  great  principles,  which  will  be  established 
in  the  next  chapter,  is  : 

Similar  areas  are  to  each  other  as  the  squares  of  any  corre- 
sponding lines. 

The  third,  which  will  be  established  in  Chapter  XIII.,  is : 

Similar  volumes  are  to  each  other  as  the  cubes  of  any  corre- 
sponding lines. 


ab 


\l 
ab 


M     a     PbN 

Fig.  114. 


CHAPTER   VI. 

70.  Theorem,  TJie  square  constructed  on  the  sum  of 
two  segments  of  a  line  equals  the  sum  of  the  squares  on  the 
two  segments  plus  twice  the  rectangle  of  „ 

the  two  segments. 

Place  the  two  segments    so   that 
MN  shall  be  their  sum. 

On  MN  as  one  side,  construct  the 
square  MH. 

At  P  erect  the  ±  PG.     On  MK  lay 
oRMD  =  a.     Draw  DE  II  to  MN.     The  student  will 
show  that: 

MDIP=a\ 

IGHE  =  h\ 
DKGI=ab, 
lENP  =  ab. 

Adding,  we  have  (a  -]-  b)'^  —  a^  -\-  b^  +  2  ab;  a  relation 
that  we  are  already  familiar  with  in  algebra.  q.  e.  d. 

71.  Theorem.  The  square  constructed  on  the  difference 
of  two  segments  of  a  line  equals  the  sum  of  the  squares  on 
the  tioo  segments  minus  ticice  the  rectangle  of  the  two 
segments. 

Place  the  two  segments  so  that  MN  shall  be  their 
difference,  MP  representing  one  segment  a,  and  PN  rep- 
resenting the  other  segment  b. 

87 


88 


ELEMENTS   OF   GEOMETRY. 


On  MN  construct  the  square  MI;  it  will  be  the  square 
on  a  —  &.  ^_  E  K 

On  a  construct  the  square 
MH.  Lay  off  DJ  and  EG  each 
equal  to  h.     Join  JG. 

JG  =  h\ 
JK=ab, 
NH=db. 


M       a     N  P 

Fig.  116. 


From  the  figure  we  see  that  (a  —  by 


a^^b^-2  ah. 

Q.  E.  D. 


72.  Theorem.  The  rectangle  having  the  sum  of  tico 
segments  for  one  side  and  the  difference  of  the  same  seg- 
ments for  an  adja^cent  side,  equals  the  difference  of  the 
squares  on  the  two  segments. 


b(a-b)p\  b" 


H 


a 


With  CE  and  CN  (as  the  sum  and  difference  respec- 
tively of  the  two  segments)  for 

adjacent    sides    construct    the    Mr- \ — ■— iJ 

rectangle  CG.  n- 

On  a  construct  the  square  CJ, 
and  on  HJ  the  square  HK. 

The  rect.  NK=  (a-b)b 

=  therect.Da 


a        BhE 
Fig.  116. 


rect.  CH  +  rect.  NK=  a^  -  b\ 

=  rect.  CH+iect.DG 
=  lect  CG  =  (a  +  b)(a  -  b). 
.-.  {a-\-b){a-b)=a^-b^; 

another  form  that  we  remember  from  algebra.         q.  e.  d. 


SQUARES   ON  SEGMENTS. 


89 


73.  Theorem.  If  squares  he  constructed  upon  the  three 
sides  of  a  right  triangle,  the  square  on  the  hypothenuse 
equals  the  sum  of  the  squares  on  the  other  two  sides. 

Let  ABC  represent  the  given  right  triangle,  right- 
angled  at  B.  On  the  side  AB  construct  the  square 
AEDB  exterior  to  the  triangle,  and  on  the  side  BC 
construct  the  square  BGKG,  overlying  a  part  of  the  tri- 
angle. 

At  the  vertex  of  the  angle  A  erect  a  perpendicular 
to  AG.  It  will  intersect  ED  in  some  point  as  at  J. 
At  G  erect  a  perpendicular  to 
AG)  and  through  J  draw  a  par- 
allel to  AG,  forming  the  rec- 
tangle AJHG. 

The  A  AEJ  is  similar  to  the 
AABG:  the  sides  of  one  be- 
ing perpendicular  to  the  sides 
of  the  other.  They  are  more 
than  similar ;  the  correspond- 
ing sides  AE  and  AB,  being  sides  of  the  same  square, 
are  equal.    Hence  the  triangles  are  equal,  and  AJ=  AG. 

The  rectangle  AJHG  is  therefore  a  square  on  the 
hypothenuse  AG. 

Draw  one  auxiliary  line,  viz.  a  ±  IIM  from  H  to  the 
line  GD. 

The  shaded  portion  of  the  square  on  AB  is  also  a  part 
of  the  square  on  AG. 

The  shaded  portion  of  the  square  on  BG  is  also  a  part 
of  the  square  on  AG. 

A  AEJ=A  GMH  (?),  and  may  be  placed  so  as  to  coin- 
cide with  it. 


Fig.  117. 


90  ELEMENTS   OF   GEOMETRY. 

A  CKN=A  HMQ  (?),  and  may  be  placed  so  as  to  coin- 
cide with  it. 

A  JDQ  =  A  AGN  (?),  and  may  be  placed  so  as  to  coin- 
cide with  it. 

Hence  all  the  parts  of  the  squares  on  AB  and  BC  have 
been  placed  so  as  to  coincide  with  the  square  on  AC 
without  repetition.  And,  furthermore,  the  square  on 
AC  has  been  completely  covered  by  the  parts  of  the 
other  two  squares.     Hence  the  theorem. 

Exercises.  —  1.  The  side  of  a  square  is  1 ;  wliat  will  the  diago- 
nal be  ? 

2.  The  side  of  a  square  is  a  ;  what  will  the  diagonal  be  ? 

3.  Show  that  the  square  on  either  of  the  perpendicular  sides  of 
a  right  triangle  equals  the  square  on  the  hypothenuse,  minus  the 
square  on  the  other  perpendicular. 

74.  Definition.  If  in  a  plane,  lines  be  drawn  through 
the  extremities  of  a  segment,  perpendicular  to  a  given 
line,  the  intercepted  portion  of  the  given  line  is  the  per- 
pendicular (orthogonal)  projection  of  the  given  segment 
on  the  given  line. 

In  the  accompanying  figure,  CD  is  the  orthogonal  pro- 
jection of  AB. 

If  through  A  and  B  par- 
allels be  drawn  obliquely  to 
the  line  CD,  the  intercept 
EF  is  an  oblique  projection 
oiAB. 

There  can  be  but  one  per- 
pendicular projection,  but  there  may  be  an  infinite  num- 
ber of  oblique  projections. 

Unless  otherwise  specified,  the  perpendicular  projec- 
tion is  the  one  meant  when  the  word  projection  is  used. 


SQUARES   ON  THE  SIDES   OF   TRIANGLES. 


91 


75.  Theorem.  If  a  triangle  he  obtuse,  the  square  con- 
structed on  the  side  opposite  the  obtuse  angle  equals  the  sum 
of  the  squares  constructed  on  the  other  two  sides,  plus  twice 
the  rectangle  formed  by  one  of  them  and  the  projection  of 
the  other  upon  it. 

Let  ABC  represent  the  obtuse 
triangle. 

Through  B  draw  BP  perpen-       a^ 
dicular  to  AG.  ^^^  ^^^ 

By  §  73,  AB"  =  AP  +  i^. 

By  §  70,  AP^  =  {AC+  CPf 

=  AO"  +  2  AC'  CP  +  CP. 

By  §  73,  PT^=C^  -  CP\ 

.■.A^  =  AB'  +  PB^  =  AC''-\-CB'-\-2AG'CP. 

Q.  E.  D. 

Exercises.  —  1.  Prove  the  theorem  by  letting  fall  a  perpen- 
dicular fi'om  A  to  the  line  BC. 

2.  Show  that  if  a  given  point  out- 
side a  circle  be  joined  to  a  point 
in  the  circumference,  M,  and  the 
point  31  be  caused  to  move  continu- 
ously about  the  circumference  from 
the  nearest  point  iV,  until  it  shall 
return  to  JV,  the  length  of  the  segment  PM  will  vary  continuously 
between  the  limits  FN  and  PF. 

76.  Theorem.  The  square  constructed  on  the  side  op- 
posite an  acute  angle  of  any  triangle  equals  the  sum  of  the 
squares  on  the  other  tivo  sides  minus  twice  the  rectangle 
formed  by  one  of  them  and  the  projection  of  the  other 
upon  it. 


Fig.  120. 


92 


ELEMENTS   OF   GEOMETRY. 


Let  ABC  in  either  figure  represent  an  oblique  triangle, 
the  interior  angle  at  C  being  acute. 


Fig.  121. 


By  §  73,  AB^  =  AK  +  PW- 

By  §71,  AP^=AG^^-PC'-2AG'PC. 

By  Ex.  3,  §  73,  PB"  =  BG^  -  PC^- 
.'.  AB"  =  aP  +  PB^  =  AO^  +  BC'-  2  AC'  PC. 

Q.  E.  D. 

Note.  —  The  accompanying  figure  is  a  convenient  one  for  help- 
ing the  memory  to  retain  the  relations  established  in  §§  73,  75,  76. 
Let  ACH  be  a  right  triangle. 


Let  ACB  be  an  obtuse  triangle 
having  ^C  and  CB  equal  to  .40 
and  CH  of  the  right  triangle. 

Let  ACK  be  an  acute  triangle       -^ 
having  AC  and  CK  equal  to  ^C 
and  CH  of  the  right  triangle. 


K  H" 


mTc 

Fig.  122. 


By  either  General  Ex.  3,  of  Chapter  II.,  or  Ex.  2,.§  75, 
AB   >AH  >AK, 
AH^  =  AC^  +  CH\ 
AB^  =  AC^BC^  +  +  2  AC-  CJ, 
AK^  =  AC^  +  CK^  -  2  AC-  MC. 


ABEAS   OF   SIMILAR   TRIANGLES.  93 

77.  Theorem.  If  in  a  right  triangle  a  line  be  drawn 
through  the  vertex  of  the  right  angle  perpendicular  to  the 
hypothenuse,  it  will  separate  the  triangle  into  two  triangles 
similar  to  the  given  one,  and  hence  similar  to  each  other. 

The  AAFB  and  ABC  have 
the  interior  angle  at  A  in  com- 
mon, and  Z  AFB=Z  ABC,  each 
being  equal  to  90°.  The  third 
angles  must  be  equal.  Hence 
the  triangles  are  similar  (§  64). 

In  the  same  way .  A  CFB  and  CBA  are  shown  to  be 
similar. 

The  A  AFB  and  CFB  having  the  angles  of  each  equal 
to  the  angles  of  ABC,  will  be  mutually  eqmangular,  and 
hence  similar.  q.  e.  d. 

(a)  By  reason  of  the  similarity  of  the  AAFB  and 
CFB,  we  have : 

AF^FB 
FB      FG' 

.-.  AF'FC=FB',  (1) 

or,  TTie  rectangle  of  the  segments  of  the  hypothenuse  equals 
the  square  of  the  perpendicular. 

(h)  By  reason  of  the  similarity  of  the  AAFB  and 
ABC,  we  have : 

AF  ^AB 

AB     AC' 

.-.  AF'AC=AB'.  (2) 

Also  from  A  CFB  and  ABC,  we  have : 

FC  ^  CB 
CB     AG' 

.'.  FG .  AC=  C^,  (3) 


94  ELEMENTS   OF   GEOMETRY. 

or,  Tlie  square  of  either  side  about  the  right  angle  equals 
the  rectangle  of  the  whole  hypothenuse  and  the  adjacent 
segment. 

(c)  Dividing  the  members  of  Eq.  2  by  those  of  Eq.  3, 
we  have : 

AF'AC     AB" 


FC'AC      CB' 
.   AF     AB" 


FC      CB 


(4) 


or,  The  ratio  of  the  segments  equals  the  ratio  of  the  squares 
of  the  corresponding  sides  about  the  right  angle. 

Adding  the  members  of  Eq.  2  to  the  members  of  Eq.  3, 
we  have : 

AF'  AC  +  FC  ■  AC  =  AB'  +  CB^. 
(AF-{-FC)AC=AB'+GB^. 

.-.  ac^  =  ab'+gb^. 

A  reproduction  of  the  relations  established  in  §  73. 

Note. — The  circumference  on  ^C  as  a  diameter  will  pass 
through  B. 

Ezercisea.  —  1.  Show  how  to  construct  a  square  equivalent  in 
area  to  a  given  rectangle. 

2.  Show  how  to  construct  a  rectangle  of  given  side,  the  area  of 
which  shall  equal  a  given  square. 

8.  Construct  a  rectangle  the  area  of  which  shall  equal  the  dif- 
ference of  two  given  squares. 

Bemark.  —  When  the  square  on  a  segment  of  a  line  equals  the 
rectangle  of  two  other  segments,  the  side  of  the  square  is  said  to 
be  a  mean  proportional  between  the  sides  of  the  rectangle. 


AREAS   OF   SIMILAR   TRIANGLES.  95 

78.   Problem.     To  find  the  relation  existing  between  the 
areas  of  similar  triangles. 


B\ 


A\ 


Fig.  124. 

Let  ABC  and  DEG  represent  the  two  triangles,  H  an 
altitude  of  ABC,  and  h  the  corresponding  altitude  of  DEG. 

Area  ABC=^BG'R. 
Area,  DEG  =  :^  EG -h. 
Dividing  member  by  member,  we  have : 
Area  ABC      BCH 


But  E^=^.    (Gen.  Ex.  4,  Chap.  V.) 


Area 

DEG 

EG'h 

BG  _ 

-^-    (G 

EG 

h 

Area 

ABC 

H' 

Area  DEG 

h' 

From  Chapter  IV.  we  know  that  the  altitudes  of  simi- 
lar triangles  are  proportional  to  any  other  corresponding 
lines  that  exist  or  may  be  drawn. 

If  B  and  b  represent  corresponding  sides, 
S^B 
h       b' 
By  squaring  both  members, 

IP^B^ 

/r         62  • 

.    Area  ABC  ^  B\ 
Area  DEG       b^ ' 


96 


ELEMENTS  OF   GEOMETRY. 


And  in  the  same  way  may  be  shown  to  have  the  same 
ratio  as  the  squares  on  any  corresponding  lines.       q.  e.  f. 

79.  Problem.     To  find  the  relation  existing  between  the 
areas  of  similar  polygons. 


Fiii.  125. 

Let  A  and  a  represent  two  corresponding  sides,  B  and 
h  other  corresponding  sides,  and  so  on  about  the  two 
figures.  The  polygons  being  similar,  the  triangles 
formed  by  drawing  corresponding  diagonals  will  be 
similar  (Gen.  Ex.  2,  Chap.  V.).  Represent  the  diag- 
onals and  areas  as  indicated  in  the  figures. 

Qi  _  Dl.    Q2  ^Dl,    Qs_^  Dl,    Qj^^p^,  ^  =  ^. 
qi        d^'    (ii        d.f    ^3         ^3=^'    q^       d/'    q^       g" 

The  second  members  of  these  equations  are  all  equal 
to  each  other. 

.  Q._A'.  Q2_A\  Q3_Al   Q4_A' 


9i       dl"     gs       d^ 


dr 


Qi 


dr 


95      di^ 


These  equations  may  be  put  in  the  form : 

QA'^q^D,'. 

QA'^^q^Di'- 

Q^.'^qsD^ 


ABEAS   OF   SIMILAR   POLYGONS. 


97 


Or,  adding  member  to  member : 

(^1+^2+^3+^4+   Q6)dl'  =  (Ql  +  g2  +  93  +  ?4  +  ^s)  A' 

Or,  Qi  +  Q2  +  Qa  +  ^4  +  Q5  ^  Dl 

gi  +  ^2  +  ^3  +  g*  +  ^5     c^i' 

If  P  and  p  represent  the  areas  of  the  respective  poly- 
gons, 

P^Dl 

P      dl 

But  the  ratio  — ^  is   the   same   as  the   ratio  of  the 

squares  of  any  other  corresponding  lines. 

The  method  of  treatment  here  employed  is  in  no  way 
dependent  upon  the  number  of  angles  of  the  polygons 
and  so  may  be  extended  to  polygons  having  any  number 
of  angles. 

Thus  we  are  not  in  danger  of  drawing  general  conclu- 
sions from  special  cases. 

Hence  the  theorem  : 

TJie  areas  of  similar  polygons  are  proportional  to  the 
squares  on  any  corresponding  lines.  Q.  e.  f. 

80.  Problem.  To  find  the  relation  that  exists  between 
the  areas  of  two  triangles  which  have  an  angle  in  each  equal. 


Fig.  126. 


Let  ABC  and  DEF  represent  the  two  triangles,  having 
the  interior  angles  at  B  and  E  equal. 


(Ex.  4,  §  CO.) 


98  ELEIHENTS   OF   GEOMETRY. 

Place  the  latter  so  that  they  will  coincide, 
Let  BHK  be  the  position  taken  by  DEF. 
Draw  the  auxiliary  line  HC 

AHBK     BK 
AHBC-BC' 

AHBC      BH 
AABC-  BA 

A  HBK •  A  HBC     BK-  BH 
AHBC'AABC~  BC'BA 

Dividing  numerator  and  denominator  of  the  first  mem- 
ber by  A  HBC,  we  have 

AHBK     BK-BH.  ^^ 
AABC-  bC-Ba'  ""^^ 

The  areas  of  two  triangles  having  an  angle  in  each 
equal,  are  proportional  to  the  rectangles  formed  by  the 
including  sides.  q.  e.  f. 

Exercises.  —  1.  Draw  the  auxiliary  line  from  J.  to  it  and 
determine  the  relation. 

2.  ABC  is  any  triangle,  DE  is 
parallel  to  AC,  DC  is  a  diagonal 
of  the  trapezoid. 

Show  that  ABDC  is  a  mean 
proportional  between  A  BDE  and 
A  BAC.  Fio.  127. 

8.  Through  the  vertex  of  a  triangle  to  draw  a  straight  line  that 
shall  separate  the  triangle  into  parts  that  shall  have  any  given 

ratio  ™. 
n 


AREAS   OF   SIMILAR   POLYGONS. 


99 


81.  Problem.  Having  given  the  three  sides  of  a  tri- 
angle, determine  the  segments  into  which  the  base  is  sepa. 
rated  by  a  perpendicular  from  the  opposite  vertex. 

Representing  the  various  segments  by  single  letters,* 
we  have: 

c2  =  a^  +  &'  -  2  by,  (§  76) 

2by  =  a'  +  b^-c', 

d'  +  W-  c" 


y  = 


(1) 

(2) 


26 
x=b  —  y. 

Equations  (1)  and  (2)  determine  the  segments. 

If  the  perpendicular  fall  without  the  triangle,  we  have ; 

c2  =  a2  +  6''  +  2  62/,         (§75) 
<^-o?-b^  =  2by, 
c'-a?-  b^ 


y  = 


2b 


x=b+y. 

The  student  will  observe  the  oneness  of  these  appar- 
ently different  results.  q.  e.  f. 

82.   Problem.     Find  an  expression  for  the  median  of  a 
triangle  in  terms  of  the  sides. 


a'=m'  +  (Jj+2,-|,i 


c^  —  m^  + 


by 


(§75) 
(§76) 


*NoTE.  —  Portions  of  geometric  figures  are  frequently  repre- 
sented by  a  single  letter,  when  misunderstanding  regarding  the 
intent  is  not  likely  to  arise  from  such  representation. 


100 


ELEMENTS   OF   GEOMETRY. 


\^  J 


(0 


=  2m2  4- 


2' 


2m?  =  a?  +  <^-^, 

o    ,     2a='  +  2c2-&2 
2m«  =  ^ ' 

,     2a2  +  2c2-62 


Fie.  180. 


m  =  ^y/2a?  +  2(^-1y'.  {u) 

Equation  (t)  expresses  the  relation  in  one  form.  (?<)  is 
simply  a  solution  of  equation  (f)  with  respect  to  m.  q.  e.  f. 

83.  Problem.  Find  an  expression  for  the  altitude  of  a 
triangle  in  terms  of  the  sides. 

Let  ABC  represent  a  triangle;  the  perpendicular,^, 
separating  the  base  into  the  segments  x  and  y. 

jP  =  a'-f,  Vb 

p'=(a  +  y){a-y),  «V 

a2  +  62  -  c^ 


Butby§81,  y  = 


26 


_  /A    Fia.  181. 


_  /^2  a6  +  g'^  +  6'  -  c^  /2  a6  -  g^*  -  6 "  -f-  c^ 
^  ~  I  26  J [  26  ; 

^  _  (g  +  6  +  c)  (g  +  6  -  c)  (g  +  c  -  6)  (6  +  c  -  g) 


METEICAL   RELATIONS. 


101 


A  more  condensed  expression  may  be  produced  by  sub- 
stituting the  perimeter  s  iov  a  -\-b  +  c.     If  a  -f  &  -f  c  =  s, 

a  +  b  —  c  =  s  —  2c, 
a-\-c  —  b  =  s  —  2b, 
b-{-c  —  a  =  s  —  2a. 

.     2_s(s-2c)(s-2b)(s-2a) 


or. 


P  =  2^Vs(s-2c)(s-26)(s-2a).   q.e.p. 


Exercise.  —  Show  that  the  formula  will  be  the  same  if  the  per- 
pendicular fall  without  the  triangle. 

84.   Problem.     Find  an  expression  for  the  area  of  a 
triangle  in  terms  of  its  sides. 

Area  =  ^  bp, 

Area  =  i &  •  ;f- -Vs{s  -2a)(s-2b)(s  - 2c),      (§  83) 
^  0 


Area  =  ^  Vs(s -2a)(s  -  2&)  (s  -  2  c). 
If  the  \  be  introduced  under  the  radical  sign, 

Area 


-^ 


s(s-2a)(s-2b)(s-2c) 


16 


Area 


=V|(|- 


.H|-.)|-o 


Exercise.  — Find  the  area  of  a  triangular  piece  of  land,  the 
three  sides  of  which  are :  32.93  chains,  48.26  chains,  and  51.48 
chains. 


102  ELEMENTS   OP   GEOMETRY. 


GENERAL    EXERCISES. 

1.  Find  expressions  in  terms  of  the  sides,  for  each  of  the 
segments  into  which  an  angle-bisector  separates  the  opposite  side 
of  a  triangle. 

2.  The  sides  of  a  triangle  are  8,  12,  and  15.  Has  the  triangle 
an  obtuse  angle  ? 

8.  Show  that  the  areas  of  two  triangles  having  an  angle  in  one 
the  supplement  of  an  angle  in  the  other  are  to  each  other  as  the 
rectangles  on  the  including  sides. 

4.  Show  that  the  sura  of  the  squares  of  the  sides  of  any  quad- 
rangle equals  the  sura  of  the  squares  of  the  diagonals  plus  four 
times  the  square  of  the  line  joining  the  middle  points  of  the 
diagonals. 

5.  Show  that  in  a  trapezoid,  the  sum  of  the  squares  of  the 
diagonals  equals  the  sum  of  the  squares  on  the  non-parallel  sides 
jp^Ms  twice  the  rectangle  on  the  parallel  sides. 

6.  Construct  a  square  equal  to  the  sum  of  two  given  squares. 

7.  Show  that /our  times  the  sum  of  the  squares  on  the  medians 
of  a  triangle  equals  three  times  the  sum  of  the  squares  on  the 
sides. 

"  8.  If  any  point  P  in  the  plane  of  a  triangle  be  joined  with  the 
three  vertices  A,  B,  C  and  the  point  of  intersection  I  of  the 
medians,  PA^  +  PFT  +  PC^  =  Ap  +  bT  +  CP  +  SPl\ 

9.  Show  how  to  draw  a  line  parallel  to  any  side  of  a  triangle  so 
as  to  bisect  the  area. 

10.  Show  how  to  draw  a  line  through  any  point  in  one  side 
of  a  triangle  so  as  to  bisect  the  area. 

11.  Show  how  to  draw  a  line  parallel  to  a  given  line  so  as  to 
bisect  the  area  of  a  given  triangle. 


CHAPTER   VII. 


85.  Theorem.  If  two  chords  intersect  within  a  circle, 
the  rectangle  on  the  segments  of  one  chord  equals  the  rec- 
tangle on  the  segments  of  the  other. 

Let  AB  and  CD  represent  two  chords  intersecting  at 
/.     Draw  the   auxiliary  lines  AC 
and  DB.     The  vertical  angles  at  / 
are  equal ; 

ZCAI=ZBDI, 

each  being   subtended  by  the  arc 
CKB. 

ZACI=ZDBI, 

each  being  subtended  by  the  arc  AD. 

.'.  AAIC  is  similar  to  A  DIB. 


Then 


—  =  —  or  AIxIB=CIx  ID. 
ID     IB 


Q.  E.  D. 


Exercise.  —  Show  how  to  construct  a  square  that  shall  be 
equivalent  to  a  given  rectangle,  the  sides  of  which  are  a  and  b.  If 
two  chords  be  so  drawn  that  the  segments  of  one 
were  adjacent  sides  of  a  rectangle  and  the  seg- 
ments of  the  other  were  equal,  by  the  theorem  we 
would  have  a  rectangle  equivalent  to  a  square. 

We  know  (§  45)  that  if  a  diameter  be  dravra 
perpendicular  to  a  chord  it  will  bisect  the  chord. 

Hence  if  a  circumference  be  constructed  with 
(a  +  &)  as  a  diameter,  and  at  the  common  extremity  of  a  and  6 

103 


Fig.  134. 


104 


ELEMENTS   OF   GEOMETRY. 


a  perpendicular  chord  be  constructed,  half  the  chord  will  be  the 
side  of  the  required  square,  since  ab  =  x^. 

The  equation  x^  =  ab  is  thus  solved  geometrically  and  exactly. 

86.  Theorem.  If  two  secants  intersect  without  the  cir- 
cumference, the  rectangle  on  the  distances  from  the  common 
point  to  the  two  intersections  with  the  circumference  in  one 
case  will  be  equal  to  the  rectangle  similarly  formed  in  the 
other. 

The  two  A  DAG  and  BEG 
are  similar.  A  C  is  common. 
Z  GDA  =  Z  GBE  (being  in- 
scribed in  the  same  segment). 
Therefore  the  third  angles  are 
equal. 

Then,  by  §  64, 

CE     GB 


Fig.  186. 


GA     GD' 


or  GE  xGD=GAx  GB.        q.  e.  d. 


Exercises.  —  1.  Having  given  a  rectangle,  construct  an  equiva- 
lent rectangle  that  shall  have  a  given  side. 

2.  K  one  secant  remain  stationary,  and  the  other  rotate  about 
the  common  point  C  until  it  become  a  tangent,  we  shall  have  a 
secant  and  a  tangent. 

Show  that  OT^  =  CA  x  CB.  ^  C. 

3.  Use  Exercise  2  to  construct  a  square 
equivalent  to  a  given  rectangle. 

4.  Construct  a  square  equivalent   to  a 
given  triangle. 

abc 


B 


Fio.  186. 


6.  x  = 


de 


Construct  z. 


be. 


Suggestion.  —  Put  x  =  -  .  — ;  construct  q  =  —,  then  construct 
_aq  d     e  e 


CHORDS   AND   TANGENTS. 


106 


87.  Problem.  Find  an  expression  for  the  bisector  of 
an  angle  of  a  triangle  in  terms  of  the  including  sides  and 
the  segments  of  the  third  side. 

Circumscribe  a  circle  about  the  triangle,  and  draw  tbe 
auxiliary  line  AK.      A.KAD  and 
CHD  are  similar. 

b  -^q      a 
c          b 
b^  -\-  bg  =  ac, 

b^  =  ac  —  bg. 
But  by  §  85,        bg  =mn. 


-.  b' 


ac  —  mn. 


If  it  be  desired  to  find  m  and  n  in  terms  of  the  sides 
a,  c,  and  d  of  the  triangle, 

m  +  71  =  a  and  —  =  — 


The  solution  of  this  pair  of  equations  will   give  m 
and  n,  in  terms  of  a,  c,  and  d.  Q.  e.  f. 

88.   Problem.     To  find  a  relation  between  the  sides  and 
the  diagonals  of  a  quadrangle  in- 
scribed in  a  circle. 

Let  a,  b,  e,  and  d  represent  the 
sides  of  the  inscribed  quadrangle, 
and  m,  n,  p,  and  q  the  segments  of 
the  diagonals. 

The  Ambq  and  dpn  are  simi- 
lar (being  mutually  equiangular).  fig.  138. 

b  _d    bd _dr         _ md? 
w,     p'  m     p  ^  p 


Then, 


(1) 


106  ELEMENTS   OF   GEOMETRY. 

Because  the  A  enq  and  apm  are  similar  (being  mutually 
equiangular), 

e     a    ae     a^  no'  /o\ 

-  =  -;  —  =  — ;  ae  = {■^) 

n     p     n      p  p 

Adding  the  members  of  equations  (1)  and  (2),  we  have 

hd  +  ae  = \p) 

P 

Equation  (3)  expresses  a  relation  between  the  sides 
and  segments  of  the  diagonals,  but  a  more  convenient 
relation  may  be  obtained  by  transforming  the  second 
member. 

By  §§75  and  76,  representing  the  perpendicular  pro- 
jection of  p  on  the  other  diagonal  by  h, 

d2=p2+  n^±2nh,  (4) 

a^  =  p^  -{-  m^  T  2  mh.  (6) 

Multiplying  both  members  of  (4)  by  m,  and  both  mem- 
bers of  (5)  by  11,  we  have 

md?  =  mp^  +  mv?  ±  2  mnh, 
no?  =  np^  4-  7n?n  T  2  mnh, 
TncP  -f  1W?  =(m  -\-  n)p^  -\-(m  -{■  n)  mn. 
But  mn=pq.  (§85) 

.'.  mcP  -f  na^  =  {m  -\-  n)p^  +{m  +  n)pq 
=  {m-\-n)(p-{-q)p. 
Substituting  in  (3),  we  have 

j,^  ■  (fn  +  n)(p  +  q)p      .      ,     ,,      ,     . 

6d  -f-  ae  =  ^= — ■ — '^^       ^'^  =  (m  +  n){p  +  5). 

Which  expressed  in  words  is : 

The  sura  of  the  rectangles  on  the  opposite  sides  equals 
the  rectangle  on  the  diagonals.  q.  e.  f. 


CHORDS   AND   TANGENTS. 


107 


89.  Problem.  Shoiv  how  to  construct  a  square  that 
shall  have  the  same  ratio  to  a  given  square  as  two  given 
straight  line  segments,  h  and  k. 


Fig.  189. 

Analysis.  —  If  the  required  square  were  known,  and  if 
sides  of  the  given  and  required  squares  were  placed  at 
right  angles  to  each  other  and  so  that  they  had  a  common 
extremity,  by  joining  their  other  extremities  a  right  tri- 
angle would  be  formed ;  and  if  from  the  vertex  of  the 
right  angle  a  perpendicular  were  drawn  to  the  hypothe- 
nuse,  it  would  separate  the  latter  into  segments  propor- 
tional to  the  squares  on  the  corresponding  sides.  [§  77  (c).] 

Construction.  —  Remembering  that  if  a  circumference 
be  constructed  with  the  hypothenuse  of  a  right  triangle 
as  its  diameter,  it  will  pass  through  the  vertex,  we 
have,  taking  m  and  n  of  convenient  length  and  so  that 

—  =  — ,  erecting  a  perpendicular  at  their  common  extrem- 
n      k 

ity  and  forming  the  A  CVA : 


If  a  side  of  the  given  square  be  less  than  a,  lay  off  VQ 


108 


ELEMENTS  OF  GEOMETRY. 


equal  to  a  side  of  the  given  square  and  draw  QR  II  to 
CA.     VR  will  be  a  side  of  the  required  square. 

VQ^     p       m      h 

If  a  side  of  the  given  square  be  greater  than  a,  lay  off 
VJ  equal  to  that  side. 

Draw  JH  II  to  CA.  VH  will  be  a  side  of  the  required 
square.  q.  e.  f. 

90.  Problem.  Show  how  to  construct  a  common  tan- 
gent  to  two  given  circumferences. 

T  s      


Fig.  140. 

Let  the  circles  whose  centres  are  C  and  C"  represent 
the  two  circles,  so  chosen  that  they  do  not  form  a  special 
case. 

Analysis.  —  If  TS  were  the  required  line  tangent  to 
the  two  circumferences  at  T  and  S,  and  if  we  should 
draw  C'S  and  CT,  they  would  both  be  perpendicular  to 
the  tangent,  and  hence  parallel  to  each  other. 

If  through  C  a  line  were  drawn  parallel  to  ST,  the 
figure  C'KTS  would  be  a  rectangle. 

If  with  C  as  a  centre,  and  CK  as  radius,  a  circle  Avere 
constructed,  C'if  would  be  tangent  to  the  constructed  cir- 
cumference at  the  point  K,  because  perpendicular  to  a 
radius  at  its  extremity.     The  radius  of  this  constructed 


CHOEDS   AND   TANGENTS. 


109 


circle  would  be  the  difference  of  the  radii  of  the  given 
circles. 

Construction.  —  With  a  radius  equal  to  the  difference 
of  the  radii  of  the  two  circles,  and  with  C  as  a  centre, 
construct  a  circle. 

Through  C"  draw  C'H,  a  tangent  to  the  newly  con- 
structed circle  (Prob.  1,  Gen'l  Ex.,  Chap.  IV.).  Draw 
CH,  intersecting  the  larger  circumference  at  J.  Through 
C"  draw  CM  parallel  to  CJ.  Each  will  be  perpen- 
dicular to  C  'H,  and  a  line  joining  the  points  M  and  J 
will  be  parallel  to  C'H, 
and  hence  perpendicular 
to  C  'M  and  CJ.  In  each 
case  being  perpendicular 
to  a  radius  at  its  ex- 
tremity, it  will  be  tan- 
gent to  each  circle,  and 
so  a  common  tangent.* 

Discussioji.  —  Since  through  any  point  outside  a  cir- 
cumference two  tangents  may  be  drawn,  we  have  a 
double  construction  as 
indicated  in  the  accom- 
panying figure. 

But  it  appears  that  a 
tangent  might  be  drawn 
that  should  pass  between 
the  two  circumferences. 

Analysis. — IfBD  were 
such  a  tangent,  and  if  we  should  draw  lines  from  C  and 


Fig.  141. 


Fro.  142. 


*  Note.  — The  student  should  construct  the  figure  as  the  read- 
ing progresses,  not  in  advance  of  it. 


110 


ELEMENTS   OF   GEOMETRY. 


Fig.  143. 


C '  to  the  points  of  tangency,  they  would  be  perpendicular 

to  the  same  line,  and  so  ^_ ^^ 

parallel.     7/"  through  C",  ^^  \      ^^ 

C  'E  were  drawn  parallel 
to  DB,  it  would  be  per- 
pendicular to  CE.  If 
then  with  C  as  a  centre 
and  a  radius  GE,  an 
auxiliary  circle  were 
drawn,    C'E   woidd    be 

tangent  to  it  at  the  point  E.    The  radius  of  the  auxiliary 
circle  would  be  the  sum  of  the  radii  of  the  given  circles. 

Construction.  —  With  C  as  a  centre  and  a  radius  equal 
to  the  sum  of  the  radii  of  the  given  circles,  construct  a 
circle.  Through  C"  draw 
a  tangent  C'Q  to  the 
auxiliary  circumference 
(Gen'l  Ex.  1,  Chap.  IV.). 
Draw  CQ,  and  through 
C"  draw  C'R  II  to  CQ. 
Each  will  be  perpen- 
dicular to  C'Q.  Draw 
WR ;  it  will  be  perpendicular  to  a  radius  at  its  extremity 
in  each  case,  and  so  tangent  to  each  circle. 

Discussion.  —  Since  from  the  point  C  two  tangents 
may  be  drawn  to  the  avixiliary  circle,  a  double  construc- 
tion may  be  had,  and  there  will  be  two  tangents  which 
pass  between  the  given  circles. 

Further  Discussion.  —  It  is  now  seen  that  there  may  be 
four  tangents  to  two  given  circumferences,  two  called 
external  tangents  and  two  internal.  If  taken  in  the  general 
position,  the  radii  remaining  constant, and  one  of  the  centres 


CHORDS   AND   TANGENTS. 


Ill 


be  moved  toward  the  other  one,  the  tangents  will  change 
the  angles  that  they  make  with  each  other,  and  when  the 


Fig.  145. 


circumferences  become  externally  tangent,  the  interior 
tangents  will  coincide  and  will  form  but  one  tangent. 

In  this  case,  which  is  a  special  one,  two  tangents  have 
become  coincident,  and  there  will  be  three  real  tangents. 


Fig.  146.  Fig.  14T. 

Further  motion  in  the  same  direction  causes  the  cir- 
cumferences to  intersect,  the  internal  tangents  are  said 
to  become  imaginary,  and  there  will  be  two  real  tangents. 

After  a  sufficient  motion  to  bring  the  circumferences 
so  that  one  shall  be  tangent  to  the  other 
internally,  the  exterior  tangents  will 
have  come  to  coincide.  We  say  that 
there  are  two  imaginary  tangents  and 
two  real,  but  coincident,  tangents. 

A  further  motion  in  the  same  direction 
will   place   one  circumference   entirely  fig.  i4S. 


112 


ELEMENTS   OF   GEOMETRY. 


within  the  other,  and  all  the  tangents  become  imaginary. 
In  other  words,  the  geometric  construc- 
tion is  impossible. 

Note.  —  The  analysis,  construction,  and  dis- 
cussion of  the  above  problem  have  been  set 
forth  thus  thoroughly,  for  the  purpose  of  ex- 
hibiting to  the  student  the  method  that  is  to  be 
pursued  in  every  such  problem.    The  student  ^ig-  149- 

should  bear  in  mind  the  fact  that  the  construction  is  the  inverse  of 
the  analysis,  and  that  each  requires  its  own  figure ;  and  that  the 
discussion  determines  the  limitations  of  the  problem. 

Henceforth  problems  of  this  character  will  not  in  the  text  be 
presented  in  such  detail,  but  the  student  is  expected  to  furnish  the 
full  detail. 

91.  Problems.  —  1.  Show  that  the  exterior  and  the  interior 
tangents  to  two  circles  intersect  on  the  line  of  centres. 


Fig.  150. 


2.  Show  that  in  either  case  the  distances  from  the  centres  to  the 
point  of  intersection  have  the  same  ratio  as  the  radii  of  the  circles.* 

8.  If  a  third  circle  be  tangent  to  two  given  circles,  the  line 
through  the  points  of  tangency  will  pass  through  the  external 
centre  of  similitude,  if  both  the  given  circles  are  externally  or  in- 
ternally tangent  to  the  third  circle. 

*  Note.  —  The  line  of  centres  is  said  to  be  divided  externally  and 
internally  in  the  same  ratio.  These  points  of  division  are  called 
the  external  and  internal  centres  of  similitude. 


CHORDS   AND   TANGENTS. 


113 


Fig.  151. 

4.  Show  that  in  the  last  figure,  PS  x  FT  =  PM  x  PQ. 

5.  Show  how  to  construct  a  circumference  tangent  to  a  given 
circumference  at  a  given  point  and  also  tangent  to  another  given 
circumference. 

6.  Show  that  if  the  centres  of  the  circles  escribed  to  a  triangle 
be  joined,  a  triangle  will  be  formed,  the  sides  of  wliich  will  pass 
through  the  vertices  of  the  given  triangle. 


Fig.  152. 

7.  Show  that  a  perpendicular  from  the  centre  of  either  escribed 
circle  to  the  line  of  centres  of  the  other  two  will  intersect  it  at  a 
vertex  of  the  given  triangle  and  will  bisect  the  interior  angle  of  the 
triangle  at  that  vertex. 


114         ELEMENTS  OF  GEOMETRY. 

8.  Show  that  the  perpendicular  mentioned  in  the  foregoing  exer- 
cise will  pass  through  the  centre  of  the  inscribed  circle. 

9.  Having  given  the  centres  of  the  escribed  circles  to  a  triangle, 
construct  the  triangle. 

10.  Apply  the  "Reductio  ad  Absurdum"  (sometimes  called 
indirect)  method  to  the  establishment  of  the  fact  that  if  three 
circles  are  tangent  to  each  other,  two  and  two,  the  common  tan- 
gents will  be  concurrent. 


Fig.  158. 

If  the  common  tangents  are  not  concurrent,  they  will  intersect 
so  as  to  form  a  triangle  as  indicated  in  the  supplementary  figure. 
By  Gen'l  Ex.  6,  Chap.  IV., 

AM=AK 
CT=  CM 


AM+BK-^  CT=  AK+BT+  CM 

But  each  term  in  the  first  member  is  less  than  a  term  in  the 
second  member. 

AM<CM 

BK<AK 
CT<BT 


AM+BK-\-  CT>CM+AK+  BT 


CHORDS   AND   TANGENTS. 


115 


Fig.  154. 


The  supposition  that  the  tangents  intersect  so  as  to  form  a  tri- 
angle is  thus  seen  to  be  an  erroneous  one. 

The  tangents  may  all  pass  through  one  point,  and  cannot  pass 
so  as  to  form  a  triangle.  Hence  they  do  all  pass  through  one 
point.  Q.  E.  D. 

11.  Show  that  if  three  circumferences  intersect,  the  common 
chords  will  all  pass  through  one  point. 

12.  It  has  been  said  that  the  three 
common  tangents  to  three  tangent 
circumferences  bisect  the  angles  of 
the  triangle  formed  by  joining  the 
points  of  tangency.     Is  it  true? 

13.  Show  how  to  separate  a  given 
segment  of  a  line  into  two  parts  such 
that  the  ratio  of  the  whole  segment  to  the  larger  part  shall  equal 
the  ratio  of  the  larger  part  to  the 
smaller  part. 

If  we  had  the  segment  separated 
as  desired,  and  if  we  let  a  represent  ( 
the  segment,  x  the  larger  part,  and 
a  —  X  the  smaller  part,  we  should 
have :  a  _     x 

X     a  —  X 
or,  a"^  —  ax  =  x^; 

a;2  +  ax  =  a^i  a;  (o  +  «)  =  a^. 

The  form  of  the  relation  suggests  that  a  secant  and  a  tangent 
to  a  circumference  might  be  used  to  express  the  relation  geo- 
metrically. 

If  PQ  represent  the  segment  a,  and  if  we  have  a  circle  tangent 
at    §,   any  secant  from  P  will  be  so 
divided  that  the  rectangle  on  the  dis- 
tances   from    P   to    the    concave    and 
convex  arcs  will  equal  a^. 

But  the  difference  of  the  distances 
from  P  to  the  concave  and  convex  arcs 
must  be  a  ;  since  one  of  the  distances  is 
to  be  x  and  the  other  is  to  he  x  +  a. 


Fio.  155. 


Fig.  166. 


116  ELEMENTS    OF   GEOMETRY. 

If  the  constructed  circle  have  for  its  diameter  a,  and  the  secant 
be  drawn  through  the  centre,  it  will  fulfil  the  required  conditions. 

Hence  at  one  extremity  of  the  given  segment,  as  at  Q,  erect  a 
perpendicular  equal  to  -•  With  -  as  a  radius  construct  a  circum- 
ference tangent  to  PQ  at  Q.  Draw  the  secant  PC.  PK  will  rep- 
resent X  or  the  larger  part  into  whicli  a  is  to  be  separated. 

If  it  be  desired  to  lay  that  part  off  on  PQ,  with  PK  as  a  radius 
and  P  as  a  centre,  construct  an  arc  that  will  intersect  PQ  at  J. 

Note.  —  When  a  segment  of  a  line  is  thus  separated  it  is  said 
to  be  divided  into  mean  and  extreme  ratio  ;  x  is  the  mean,  and  the 
extremes  are  a  and  a  —  x.  When  written  in  extended  form  it  is : 
a  '.x::x:a  —  x. 


CHAPTER   VIII. 


Definition.  A  regular  polygon  is  one  the  angles  of 
■which  are  equal  to  each  other  and  the  sides  are  equal 
to  each  other. 

92.  Theokem.  a  circumference  may  be  circumscribed 
about  any  regular  polygon. 

Let  the  accompanying  figure  represent  a  regular  poly- 
gon. The  bisectors  of  two  adjacent  angles  will  meet 
at  some  point,  as  C,  forming  the  isosceles  A  CAB. 

If  C  be  joined  with  D, 

ACBD=ACBA. 

.-.A  CBD  is  isosceles,  CB=CD, 
and  the  angle  of  the  polygon  at 
D  is  bisected. 

If  C  be  joined  with  the  other 
vertices,  triangles  will  be  formed 
each  equal  to  CAB. 

Therefore  the  distances  from 
the  point  of  intersection  of  the  bisectors  of  a  pair  of 
adjacent  angles  from  all  the  vertices,  is  the  same,  and  if, 
with  this  point  as  a  centre  and  a  radius  equal  to  the  dis- 
tance from  this  point  to  any  vertex,  a  circumference  be 
described,  it  will  pass  through  each  vertex.  q.  e.  d. 

Kemakks.  —  The  centre  of  the  circumscribed  circle,  which  is 
also  the  centre  of  the  inscribed  circle,  is  called  the  centre  of  the 
polygon. 

117 


118 


ELEMENTS   OF   GEOMETRY. 


The  radius  of  the  circle  circumscribed  about  a  regular  polygon 
is  called  the  radios  of  the  polygon. 

The  radius  of  the  circle  inscribed  in  a  regular  polygon  is  called 
the  apothem. 

Ezercises.  —  1.  Show  that  perpendiculars  from  C  to  the  sides 
of  the  polygon  will  all  be  equal,  and  that  a  circumference  may  be 
inscribed  within  a  regular  polygon. 

2.  Show  that  an  inscribed  equilateral  polygon  will  be  equi- 
angular, and  hence  regular. 

8.  Show  that  an  inscribed  equiangular  polygon  may  not  be 
regular.     Discuss. 

4.  Show  that  an  equiangular  polygon  circumscribed  about  a 
circumference  is  regular. 

6.  Show  that  regular  polygons  of  the  same  number  of  sides  are 
similar  figures. 

93.  Theorem.  If  a  regular  polygon  he  inscribed  in  a 
circumference  and  then  tangents  he  drawn  parallel  to  the 
sides  of  the  inscribed  polygon,  a  circumscribed  polygon  vxill 
he  formed  that  will  he  similar  to  the  inscribed  polygon. 

Let  DA,  AB,  and  BE  represent  three  adjacent  sides  of 
the  regular  inscribed  polygon.  Let  CF,  CG,  and  CH  be 
the  perpendicular  bisec- 
tors of  these  sides.  They 
will  also  bisect  the  sub- 
tended arcs  at  0,  M,  and  J. 

If  at  these  points,  tan- 
gents be  drawn,  they  will 
be  parallel  to  the  sides  of 
the  inscribed  polygon. 


Fig.  168. 


ZONM=ZFAG, 
ZMKJ=ZGBH, 

etc.  etc. 

The  angles  of  the  circumscribed  polygon  equal  those 


ENSCHIBED   AND   CIRCUMSCRIBED   POLYGOKS.       119 


of  the  inscribed,  since  the  sides  are  parallel  and  extend 
in  the  same  direction. 

Hence  (Ex.  4,  §  92)  the  circumscribed  polygon  is  regu- 
lar. But  each  side  of  the  latter  corresponds  to  a  side  of 
the  inscribed  polygon.  Therefore  they  have  the  same 
number  of  sides,  and  being  regular,  are  similar.       q.  e.  d. 

94.  Theorem.  If  at  the  vertices  of  a  regular  inscribed 
polygon,  tangents  he  drawn,  a  circumscribed  polygon  will  be 
formed  wJiich  will  be  similar  to  the  inscribed  one. 

Let  A,  B,  C,  D,  etc.,  represent  the  vertices  of  the  in- 
scribed polygon,  at  which  tan- 
gents are  drawn,  intersecting  at 
E,  F,  G,  etc. 

The  A  ABE,  BCF,  and  CDG 
are  isosceles  and  are  equal  to 
each  other  (each  having  a  side 
and  two  adjacent  angles  in  each 
equal). 

Therefore  the  angles  of  the 
circumscribed  polygon  are  all  equal  to  each  other. 

Each  side  of  the  circumscribed  polygon,  being  the  sum 
of  two  segments  (each  of  which  is  equal  to  any  one  seg- 
ment, as  BF),  are  equal  to  each  other. 

Hence  the  polygon  constructed  fulfils  the  requirements 
for  regularity,  and  having  the  same  number  of  sides  as 
the  inscribed  polygon  will  be  similar  to  it.  q.  e.  d. 

Exercises.  —  1.  Show  how  to  cut  a  square  piece  of  board  so 
as  to  get  a  regular  octagon. 

2.  Show  that  if  chords  be  drawn  from  the  vertices 
of  a  regular  inscribed  polygon  to  the  middle  points 
of  the  subtended  arcs,  a  new  regular  inscribed  poly- 
gon of  double  the  number  of  sides  will  be  formed,  Fio-  160. 


Fig.  159. 


120  ELEMENTS   OF   GEOMETRY. 

3.  Show  that  if  the  alternate  vertices  of  a  regular  polygon  of 
an  even  number  of  sides  be  joined,  a  regular  polygon  of  half  the 
number  of  sides  will  be  formed. 

4.  Construct  regular  polygons  of  3,  4,  6,  8,  12,  and  16  sides. 

6.  Show  that  the  area  of  a  regular  polygon  equals  half  the  rec- 
tangle on  its  perimeter  and  apothem. 

6.  Show  that  the  area  of  any  polygon  circumscribed  about  a 
circle  equals  half  the  rectangle  on  its  perimeter  and  the  radius  of 
the  circle. 

95.  Theorem.  If  a  regular  polygon  be  circumscribed 
about  a  circle,  lines  be  drawn  from  the  centre  to  the  vertices, 
and  tangents  be  drawn  at  the  points  where  these  lines  inter- 
sect the  circumference,  a  regular  circumscnbed  polygon  of 
double  the  number  of  sides  will  be  formed. 


Fig.  161. 

Let  HT  represent  a  side  of  a  circumscribed  regular 
polygon,  G  and  K,  points  at  which  tangents  are  drawn  as 
required.  MK=KD=DJ, 

BJ=  BG, 
GB  =  DK 
(because  AHGB  =A  DKT). 
.'.  MK=  KD=DJ=JB  =BG=  GN. 
..MD  =  DB=BN. 


VARIABLE   AND   LIMIT.  121 

The  same  may  be  shown  for  the  other  sides  of  the  new 
figure.     Therefore  the  new  figure  will  be  equilateral. 

'.'A  TKM,  TKD,  HGB,  and  HGN  are  equal  to  each 
other,  Z  TMK=Z  TDK=Z  HBG  =/.  IING. 

The  supplements  of  these  several  equal  angles  are 
interior  angles  of  the  polygon.  Therefore  the  interior 
angles  of  the  new  polygon  are  all  equal. 

Hence  the  new  polygon,  which  will  have  double  the 
number  of  sides  of  the  original  polygon,  will  be  both 
equilateral  and  equiangular:  which  are  the  conditions 
of  regularity.  q.  e.  d. 

VARIABLE  AND  LIMIT. 

96.  Definitions.  A  variable  is  a  changing  quantity; 
such  as  the  time  that  has  elapsed  from  some  given  epoch 
to  the  present,  which  is  always  changing. 

That  which  represents  the  height  of  the  ocean  tide,  the 
chord  of  a  circle  moving  parallel  to  itself,  or  the  speed 
of  a  bullet  through  the  air,  are  further  examples. 

As  a  further  illustration  let  perpendicular  diameters 
be  drawn  in  a  circle.     The  point  gen- 
erating  the    circumference    will    be   at 
changing  distances  from  these  diameters. 
If  X  represent  the   distance  of  P  (any 
given  point  on  the  circumference)  from 
one   diameter,  and  y  represent  the  dis- 
tance from  the  other  diameter,  x  and  y  ^^' 
will  be  variables.     Incidentally  they  are  so  related  that 
a^+y^=:r^. 

If  there  be  a  fixed  quantity  toward  which  a  variable 
approaches,  that  fixed  quantity  is  called  a  limit.    The  law 


122  ELEMENTS   OF  GEOMETRY. 

governing  a  variable  will  determine  whether  it  has  a 
limit  or  not. 

To  illustrate:  The  repeating  decimal  .6666...  will 
approach  the  fixed  quantity  f ;  the  larger  the  number  of 
terms  considered,  the  nearer  will  the  repeating  decimal 
come  to  ^.  In  this  case  the  limit  can  be  seen  and 
named,  although  never  reached  by  increasing  the  number 
of  terms  in  the  decimal. 

If  a  secant  AB  rotate  about  the  point  ^  as  a  pivot,  B 
will  approach  A.  When  B  shall  coin- 
cide with  A,  the  line  will  have  ceased 
to  be  a  secant  and  will  have  become  a 
tangent.  If  the  point  B,  which  is  at 
a  variable  distance  from  A,  shall  con- 
tinue to  move  along  the  circumference 
after  having  coincided  with  A,  the 
line  will  again  become  a  secant.  The 
tangent  is  said  to  be  the  limit  toward  which  the  secant 
approaches  as  the  second  point  of  intersection  approaches 
the  first. 

In  this  case  the  secant  reaches  its  limit. 

The  series  l+y+i  +  i  +  TV+'*'  ^^  ^^^  ^^^  limit 
the  number  2,  which  we  can  see  and  can  name,  although 
it  cannot  be  reached  by  any  increase  in  the  number  of 
terms. 

Let  this  problem  be  proposed :  At  what  time  between 
2  and  3  o'clock  will  the  hour  and  minute  hands  of  the 
clock  be  together  ? 

At  2  o'clock  10  minute  spaces  separate  the  hands. 
When  the  minute  hand  shall  have  reached  the  position 
primarily  occupied  by  the  hour  hand,  it  will  be  behind 


VARIABLE   AND   LEVIIT.  123 

the  hour  hand  -^  of  the  distance  that  first  separated 
them.  When  the  minute  hand  shall  have  reached  the 
position  occupied  by  the  hour  hand  at  the  last  account- 
ing, the  hour  hand  will  be  in  advance  -^  of  the  distance 
that  previously  separated  them. 

Then  the  number  of  minute  spaces  to  be  passed  over 
by  the  minute  hand  will  be  : 

10      10        10         10         10 

"•"  12  "^  144  "^  1728  "*■  (12)*  +  (12)^  +  *  *  * 

That  this  series  approaches  a  definite  limit  which  can 
be  named,  we  can  determine  by  finding  in  another  way 
the  number  of  minute  spaces  passed  over. 

Let  X  represent  the  number  of  minutes 
after  2  that  the  hands  are  together  (for 
they  will  be  together).     Then : 

(X  —  10)  12  =  X,  Fig.  164. 

1205-120=3?, 

11  a;  =  120,  x  =  10\^. 

Therefore  the  series  above  written  approaches  lOif  as 
its  limit. 

If  0  represent  one  angle  of  a  regular  polygon  of  n  sides, 

e  =  180°  -  5^ 

n 

360° 

As  11  increases, will  decrease,  and  6  approaches 

n 

180°  as  its  limit. 

If  we  have  a  series  of  numerical  terms  represented  by 
a  +  hh  +  ch? -\-(W -{- eh^ -\- '•• ,  the  limit  toward  which 
this  series  will  approach  as  we  diminish  the  value  of  h 
will  be  a. 


124         ELEMENTS  OF  GEOMETRY. 

A  limit  may  be  of  the  same  kind  as  the  magnitude 
which  approaches  it,  or  it  may  be  of  a  different  kind. 

Depending  upon  the  natui-e  of  variable  and  limit,  a 
limit  may  be  reached,  or  it  may  not  be  reached. 

There  are  two  axioms  relating  to  variables  and  limits 
which  are  of  great  importance. 

Axiom  I.  If  the  ratio  of  two  variable  quantities  which 
approach  limits  always  equals  a  fixed  quantity,  as  the  vari- 
ables approach  their  limits,  the  limits  (or  limiting  values) 
will  have  the  same  ratio. 

Illustration. ^  =  -j  — -  =  -'}  — r  =  -,  etc. 

Yi     b     Y2     b     Ys     b 

If  Xu  Yi,  Xi,  Y2,  etc.,  represent  variables  that  at  each 

stage  have  the  same  ratio,  ->  and  these  variables  approach 

b 

X  and  Y  as  their  limits,  :^  =  r* 

Axiom  II.  If  two  ratios,  the  terms  of  which  are  comr 
posed  of  variables  apjtroaching  limits,  are  always  equal  to 
each  other,  as  the  limits  are  approached,  they  will  be  equal 
to  each  other  at  the  limits. 

Illustration.  —  Wi,  Xi,  Yi,  Zi,  W23  X^,  etc.,  represent 
limit-approaching  variables,  and  at  different  halting- 
places  of  investigation  we  have 

^=Il,    ^  =  1^,   ^=^3,   etc. 
Xi      Zi     X2      Z2     -X3      Z^ 

Then  if  W,  X,  Y,  and  Z  represent  the  limits  approached, 

X    z 


VARIABLE   AND   LIMIT.  125 

Exercises.  —  Using  the  Limits  Axioms^  — 

1.  Show  that  if  a  line  be  drawn  parallel  to  a  side  of  a  triangle, 
it  divides  the  other  sides  proportionally.  (Suggestion :  Approach 
the  incommensurable  limits  through  successive  commensurable 
ratios.) 

2.  Show  that  in  the  same  or  equal  circles,  angles  at  the  centre 
are  proportional  to  the  intercepted  arcs. 

3.  Show  that  the  areas  of  two  rectangles  are  to  each  other  as 
the  products  of  base  and  altitude. 

97.  Theorem.  If  a  regular  polygon  he  circumscribed 
about  a  circle,  and  a  regular  polygon  of  double  the  number 
of  sides  be  constructed  (as  indicated  in  §  95),  the  latter  ivill 
be  of  less  area  than  the  former,  and  of  less  perimeter. 

Using  the  figure  of  §  95,  we  see  that  if  n  represent  the 
number  of  sides,  when  the  number  of  sides  is  made  2n, 
that  n  isosceles  triangles  (like  DTM)  will  have  been 
eliminated,  and  nothing  having  been  added,  the  area  will 
have  been  diminished. 

Furthermore,  with  each  elimination  of  an  isosceles  tri- 
angle, the  base  takes  the  place  of  the  other  two  sides  as 
a  part  of  the  perimeter,  and  therefore  the  perimeter  is 
diminished. 

Exercises.  —  1.  Show  that  the  circumferences  of  two  circles 
have  the  same  ratio  as  their  diameters. 

2.  Show  that  the  areas  of  two  circles  have  the  same  ratio  as  the 
squares  of  their  radii. 

Note.  —  The  student  will  observe  that  at  each  doubling  of  the 
number  of  sides  of  the  circumscribed  polygon,  the  area  will  be 
diminished.  The  area  of  the  circle  is  the  limit  toward  which  the 
area  of  the  circumscribed  polygon  approaches  as  the  number  of 
sides  is  increased  by  doubling. 


126 


ELEMENTS   OF   GEOMETRY. 


Fig.  166. 


98.  Theorem.  If  a  regular  polygon  he  inscribed  in  a 
circle,  and  a  regular  polygon  of  double  the  number  of  sides 
be  constructed  (as  indicated  in  Ex.  2,  §  94),  the  latter  loill 
be  of  larger  area  than  the  former  and  of  greater  perimeter. 

Let  AB  represent  a  side  of  a  regular  inscribed  polygon, 
and  AD  and  DB  represent  sides  of  an         .       T) 
inscribed  polygon  of  double  the  number 
of  sides. 

If  n  represent  the  number  of  sides  in 
the  given  polygon,  when  the  number  of 
sides  is  doubled  there  will  be  added  to 
the  area  of  the  polygon  n  isosceles  tri- 
angles ;  and  for  each  side  of  the  poly- 
gon of  n  sides  there  will  be  substituted  AD+DB  for 
AB,  etc.     Hence  the  theorem.  q.  e.  d. 

Note.  —  The  student  will  observe  that  at  each  doubling  of  the 
number  of  sides  of  the  Inscribed  polygon  the  area  will  be  increased 
no  matter  how  many  times  the  number  of  sides  be  doubled. 

But  the  area  of  the  polygon  will  always  be  less  than  the  area  of 
the  circle,  toward  which  we  approach  as  the  process  goes  on. 

The  circle  is  the  limit  toward  which  the  inscribed  polygon 
approaches  as  we  increase  the  number  of  sides. 

99.  Pkoblem.  To  determine  the  law  by  which  the  areas 
of  inscribed  and  circumscribed  regular  polygons  approach 
the  area  of  the  circle  as  the  sides  are  doubled  in  number. 

Let  DB  and  MN  represent  sides  of  ^Af  H^  K^J    Ny 
an  inscribed  and  a  circumscribed  poly- 
gon of  n  sides. 

DK  will  represent  a  side  of  the 
inscribed  polygon  of  2  n  sides,  and  HJ 
will  represent  a  side  of  the  circum- 
scribed polygon  of  2  71  sides.  fio.  lee. 


VARIABLE   AND   LIMIT.  127 

If  a  represent  the  area  of  the  inscribed  polygon,  and 
A  the  area  of  the  circumscribed  polygon  of  n  sides, 

ADCB  =  %     ADCQ=^^f^  =  ^, 

n  2  2n 

n  A  In 

.    g  ^  A  DGQ 
"A     A  MCK 

2  n  (A  DGK)  equals  the  area  of  the  inscribed  polygon 
of  2  n  sides,  and  2n{A  HJC),  the  area  of  the  circum- 
scribed polygon  of  2w  sides.  If  a'  and  A'  represent 
these  areas  respectively, 

ct'  ^  2  n  (A  DCK)  ^  A  DGK 
A'      2n{AHGJ)      AHGJ 

By  §  80,  the  A  DGK  is  a  mean  proportional  to  the 
A  DGQ  and  MGK.  Each  multiplied  by  2  w  would  pro- 
duce the  entire  areas  of  the  polygons  of  which  they  are 
parts. 

.'.a'  =  VaA,  or  —  =  — ; — >  (1) 

AHGK^KH 
A  MGK     KM 
But  4.n'AHGK  =  A' 

and  2n-AMGK=A, 

or  4.11- A  MGK  =2  A. 

.    A'  ^A  HGK^  KH 
"  2A     A  MGK     KM 

^,^KH(2A)^  2KH-A 
KM         KH+HM 


128 


ELEMENTS    OF   GEOIMIETKV 


Taking  the  reciprocals  of  the  last  result,  we  have 

^'       2KHA       '^\    KH'A 

^      1/    KH     ,     HM 

A 


But 


or 


a 
A 


KH'  A 


+ 


KH 


a)     2^ 


HM 
^     KH- 


KH 


CM       CM      HM 


=  4=^    §  69,  Prob.  6. 


KH 

hm' 

HM 


A  or  Va 


HM 


KH-A     KHVaVA     (KH 


1^ 

A' 


KH 
HM 

1 


V3. 


HM 
1/1 


..r7. 


aVA 


The  reciprocals  of  the  areas  are : 

a    ^    -y/oA    ^\A 

If  then  we  know  the  reciprocals  of  the  areas  of 
inscribed  and  circumscribed  polygons  of  n  sides,  the 
reciprocals  of  those  of  2  n  sides  may  be  determined  by 
the  above  established  relations. 

100.  A  Numerical  Computation.  —  If  in  a  circle  whose 
radius  is  unity  we  inscribe  a  square, 
and  then  circumscribe  a  square 
about  it,  we  would  have  for  the 
area  of  the  inscribed  square,  a  =  2; 
and  for  the  area  of  the  circum- 
scribed square,  A  =  4:-,  for  the  area 
of  the  inscribed  octagon,  a'  =  V8.  ^^^  ^^^ 


AREAS  OP  REGULAK  POLYGONS.       129 

1  =1=.5.  i  =  l=.26. 
a        2  A     4: 


-  =  V.5  X  .25  =  .3o35o33907  ... 

~-  =  i(.25  +  .3535533907)  =  .3017766953 
■^1 


,3266407412... 


^  =l(~  +-)    =  .3142087182 , 

A2      2\Ai     a  J 

1=^14   =.32 


3203644309... 


^  =  ^  (^4-  +  -^  =  .3172865745  . . . 

A3      2  \A.2     OsJ 

A  =J'L.1_  =.3188217885... 

-^  =Jf-4  +  -')  =.3180541815... 


.3184377537 


1  =  l.JL 

4-  =  ^  f  ^  +  -^  =  .3182459676  . . . 

A,      2\Ai     a  J 


1  _   ^71 

26  ^'a4      -^5 

Ae      2\A,     a. 


,3183418462... 


^  =l(~+-]    =  .3182939069 . . . 


,3183178756... 


aj        Vog    A^ 


130  ELEMENTS  OF  GEOMETRY. 


=  -  /^—  + 1")  =  .3183058912  . . . 


1 

1 

As 

_  /I   1 

2U7    V 

.3183118833 ... 
.3183088872... 

1 
1 

/I  1 

2  Us       V 

.3183103853... 
.3183096362... 

Oio 
1 

•^10 

2VA     Oio/ 

.3183100107... 
.3183098234... 

1 

_     /I   .    1 
~2Uo'^aJ~ 

.3183099170 ... 
.3183098702 ... 

—  =  -J——  =  .3183098936  ... 

fltl2     ^  (111   -^11 

-^  =  1(^4-  +  — 1  =  -3183098819 ... 
A^     2  \^ii  012/ 

—  =  -J—  .  —  =  .3183098877  . . . 

Oqg  '  CI12    ^12 

4-  =  ^/"-^  +  —  V  .3183098848 ... 
Aa     2\An     a^J 


AREA   OF   A   CIRCLE.  131 

ai3  and  Ajs  are  respectively  the  areas  of  regular  poly- 
gons of  32768  sides,  inscribed  within  and  circumscribed 
about  a  circle ;  the  square  on  the  radius  being  the  unit. 
1 


.318309887 


3.1415926386... 


Aio  = =  3.1415926672  ... 

''     .3183098848... 

The  area  of  the  circle  lies  between  the  areas  of  a„  and 
A„,  and  is  nearer  to  either  of  them  than  they  are  to  each 
other. 

The  same  may  be  said  of  ai  and  A-^,  a^  and  A^,  Og  and 
As,  etc.,  as  far  as  the  computations  may  be  continued. 

In  our  investigation  we  have  carried  the  approximation 
to  Oia  and  A^^,  and  find  that  they  are  the  same  for  seven 
decimal  places.  The  area  of  the  circle  will  be  the  same 
for  seven  decimal  places. 

Letting  K  represent  the  area  of  a  circle  and  B  its 
radius,  we  will  have : 

7^  =  (3.1415926...)  (i?^), 

or  -^2  =  3.1415926... 

The  area  of  a  circle  is  an  exact  thing  and  the  square 
on  the  radius  is  an  exact  thing.  The  ratio  of  the  two, 
by  common  consent,  is  represented  by  the  Greek  letter 
(tt)  and  is  an  incommensurable  number  to  which  we  may 
approach  as  near  as  time  and  patience  will  allow,  but 
which  we  can  never  express  in  integers  or  their  fractional 
parts. 

Note.  —  The  quadrature  of  a  circle  (^■.e.,  the  determination  of 
its  area  in  terms  of  any  given  square)  has  demanded  the  attention 
of  students  of  mathematics  for  4000  years  or  more,  and  has  had 
expended  upon  it  a  vast  amount  of  time. 


132 


ELEMENTS   OF   GEOMETRY. 


The  earliest  statements  that  we  have  give  the  ratio  as  3,  later 
(1700  B.C.)  as  about  3.16,  still  later  (220  b.c.)  as  3^.  The  most 
remarkable  ratio  of  whole  numbers  that  approximate  to  ir  (1600 
A.D.)  is  fff,  which  when  expressed  decimally  agrees  with  w, 
expressed  decimally,  to  the  sixth  decimal  place. 

f  f-|  may  be  easily  remembered  by  noting  the  fact  that  if  we 
write  the  first  three  odd  numbers  twice  each,  and  then  divide  the 
last  three  by  the  first  three,  thus,  113)355,  we  have  the  ratio. 

If  3.1415926  be  expressed  as  a  continual  fraction,  it  gives  : 
1 


T  =  3  +  -         1 

7+—      1 
15  + - 
1  + 


190  + 


For  which  the  first  three  succeeding  convergents  are : 

C8  =  3^A  =  Hf. 

The  terms  of  the  series,  1-,  J,  V^-^,  etc.,  are  (beginning with  the 
fourth  term)  alternately  the  arithmetical  and  the  geometrical  means 
of  the  two  preceding  terms ;  and  is  known  as  Schwab's  series. 

Bemark.  —  Because  of  the  fact  that  we  are  unable  to  express 
the  area  of  a  circle  in  terms  of  the  square  on  the  radius  exactly, 
it  does  not  follow  that  other  areas  bounded  by  curved  lines,  or  by 
a  combination  of  curved  and  straight  lines,  cannot  be  expressed 
exactly  in  terms  of  a  square  on  some  given  line,  or  a  rectangle,  or 
a  parallelogram,  which  may  be  converted  into  a  square. 

Illustration.  —  The  locus  of  a  point 
which  moves  in  a  plane  so  that  the  ratio 
of  its  distances  from  a  fixed  point  and 
from  a  fixed  line  equals  1  generates 
what  is  called  a  parabola. 

If  through  any  point  (P)  a  line  (PPi) 
parallel  to  the  given  line  {MMi)  be 
drawn,  the  area  included  between  the 
curve  and  PPi  will  be  exactly  equal  to 
f  the  rectangle  P^^iPi- 

This  fact  will  be  established  in  §  163.  fiq.  les. 


M 

0                   P 

^ 

M, 

\^ 

Qi              Pi 

AREA   OF   A   CIRCLE.  133 

101.  By  Exs.  5  and  6,  §  94,  we  see  that  the  area  of 
any  regular  polygon  equals  the  rectangle  on  the  half- 
perimeter  and  the  apothem,  or  frequently  expressed  as 
half  the  product  of  perimeter  and  apothem.  The  same 
holds  true  no  matter  how  many  sides  the  polygon  may 
have. 

If  successive  polygons,  doubling  in  number  of  sides,  be 
inscribed  within  a  circumference,  the  perimeter  will  con- 
tinually increase,  approaching  the  circumference  of  the 
circle  as  its  limit,  and  the  apothem  will  increase,  approach- 
ing the  radius  of  the  circle  as  its  limit. 

If  successive  polygons,  doubling  in  number  of  sides, 
be  circumscribed  about  a  circumference,  the  perimeter 
will  continually  decrease,  but  the  apothem  will  remain 
constant. 

In  each  case  the  area  equals  the  half-product  of  perim- 
eter and  apothem. 

When  we  have  reached  32768  sides  for  each,  the  areas 
of  the  circumscribed  and  inscribed  regular  polygons  (in 
terms  of  the  square  on  the  radius)  agree  to  seven  deci- 
mal places.  The  area  of  the  circle  lies  between  the  two 
areas,  and  the  circumference  of  the  circle  is  the  common 
limit  upon  which  the  perimeters  of  the  inscribed  and  cir- 
cumscribed polygons  converge ;  we  therefore  say  that 
the  area  of  a  circle  equals  half  the  product  of  its  circum- 
ference and  radius.  If  A  represent  the  area  of  a  circle, 
C  its  circumference,  and  H  its  radius, 

A  =  ^C-E. 

Expressed  in  general  language : 

TJie  area  of  a  circle  equals  the  half  product  of  the  cir- 
cumference and  the  radius. 


134  ELEMENTS   OF   GEOMETRY. 

102.  From  §  100, 

A  =  irR-.     But  A  =  \CR. 

.•.^CB  =  7rB^,0v\C=irR. 

C  =2 ttB,  (form  most  used) 

C 
or  C  =  ttD,  or  —  =  tt. 

Hence  it  may  also  be  described  as  the  ratio  of  the  cir- 
cumference to  the  diameter  of  a  circle,  as  well  as  the 
ratio  of  the  area  of  a  circle  to  the  square  on  its  radius. 

103.  Theorem.  The  perimeter  of  any  polygon  inscribed 
within  a  circle  is  less  than  the  circumference,  and  the 
perimeter  of  any  polygon  circumscribed  about  a  cirde  is 
GREATER  than  the  circumference  of  the  circle. 

(a)  Each  side  of  the  inscribed  polygon  is  a  straight 
line,  joining  two    points,   hence  will    be 

shorter  than  the  arc  which  it  subtends. 
The  sum  of  these  chords  —  which  is  the 
perimeter  of  the  polygon  —  will  therefore 
be  less  than  the  sum  of  the  subtended  arcs 
—  which  is  the  circumference  of  the  circle.        ^^^  ^g^ 

Q.E.  D. 

(b)  By  Ex.  6,  §  94,  the  area  of  a  circumscribed  polygon 
equals  the  half-product  of  the  perimeter  and  the  radius 
of  the  circle. 

A'^^PE. 
By  §  101,  the  area  of  the  circle  equals  the  half-product 
of  its  circumference  and  radius. 
A=\  CR. 
But  the  area  of  the  circumscribed  polygon  is  greater 
than  the  area  of  the  circle. 

..  ^PR>\CE,01  P>C.  Q.E.D. 


AREA   OF   A   CIRCI.E. 


136 


Fig.  ITO. 


Exercises.  —  1.    Show  that  the  area  of  any  sector  equals  half 
the  product  of  the  included  arc  and  the  radius. 

2.  Find  the  areas  of  sectors  of  30°,  of  45°,  of  120°,  of  300°, 
of  390°,  of  480°,  and  of  540°. 

3.  The  areas  of  circles  are  to  each  other  as  the  squares  of  their 
radii,  or  the  squares  of  their  diameters,  or  the 

squares  of  their  circumferences,  or  the  squares 
of  any  corresponding  lines. 

4.  If  at  the  extremities  of  an  arc  AB,  which 
determines  a  sector  of  a  circle,  tangents  be  drawn 
intersecting  at  I,  AI  +  IB  will  be  greater  than 
the  arc  AB. 

5.  Show  that  if  two  circumferences 
intersect,  the  subtended  arcs  on  each 
side  of  the  common  chord  will  be  un- 
equal, and  the  enveloping  one  will  be 
the  longer. 

Show  how  to  determine  by  tangents 
at  J^  or  5  which  curve  will  be  the  en- 
veloping one.  Discuss  the  problem 
thoroughly. 

104.    Problems.  — 1.   To  inscribe  a  regular  decagon  in  a  circle. 

Analysis.  —  If  AB  were  the  side  of  a  regular  inscribed  decagon, 

and  if  we  should  join  the  vertices  of  the 

polygon  to  the  centre,  we  should  have  ten 

triangles  formed,  each  equal  to  ACB. 

ZACB  =  S6°, 

jL  cab  =  72°, 

Z  CBA  =  72°. 

If   the  Z  CBA  be  bisected  by  the  line 

BD,  we  would  have  the  /i^  CBD  and  ABD 

isosceles  as  well  as  the  A  ACB. 

The  A  ACB  and  ABD  being  mutually  equiangular,  are  similar. 
.    CA^BD 
' ' AB      DA 


Fig.  171. 


136 


ELEMENTS   OF   GEOMETRY. 


Fig.  173. 


But  AB  =  BD=  CD. 

■■  CD      DA 
Which  means  that  the  radius  will  be  separated  into  mean  and 
extreme  ratio. 

Construction. — By  Prob.  13,  §91,  separate  the  radius  of  the 
circle  in  which  the  decagon  is  to  be  in- 
scribed, into  parts  such  that  the  ratio  of 
the  whole  radius  to  the  larger  part  equals 
the  ratio  of  the  larger  part  to  the  smaller. 
CK^  CQ 
CQ      QK  ' 
From  K  lay  off   chords  equal  to  CQ; 
they  will  form  the  sides  of  the  required  decagon. 

2.  Show  that  if  the  alternate  vertices  of  a  decagon  be  joined,  a 
regular  pentagon  will  be  formed. 

3.  Making  use  of  §  88,  show  how  to  find  the  diagonals  of  a  reg- 
ular pentagon,  having  given  the  length  of  one  side. 

4.  Show  how  a  regular  hexagon  and  a  regular  decagon  may  be 
used  to  inscribe  a  regular  polygon  of  16  sides,  and  hence  one  of 
80  sides. 

6.   Show  that  the  square  on  the  side  of  a  regular  inscribed  pen- 
tagon equals  the  sum  of  the  squares  on  a 
side  of  the  regular  inscribed  decagon  and 
on  the  radius  of  the  circle. 

6.  Having  given  the  side  of  a  regular 
decagon,  find  the  corresponding  radius. 

7.  Find  the  area  of  a  regular  inscribed 
dodecagon  in  terms  of  the  square  on  the 
radius.  Fzo.  174. 

8.  Show  how  to  make  a  five-pointed  star  and  determine  the  sum 
of  the  angles  at  the  points. 


SOME   PROBLEMS   IN   PLANE 
GEOMETRY. 

(Note.  — Do  not  neglect  the  analysis.) 

Show  that : 

1.    The  three  altitudes  of  a  triangle  are  concurrent  lines. 


Fig.  1T5. 

Suggestion.  —  Through  the  vertices  of  the  given  triangle  draw 
lines  parallel  to  the  opposite  sides. 

2.  A  quadrangle,  the  opposite  sides  of  which  are  equal,  is  a 
parallelogram. 

8.  Find  the  area  of  a  triangle  the  sides  of  which  are  12,  16, 
and  20.  Find  the  length  of  the  median  from  the  largest  angle. 
Determine  the  radii  of  the  circumscribed  and  inscribed  circles. 

4.  If  from  any  point  in  the  base  of  an  isosceles  triangle  lines 
be  drawn  parallel  to  the  equal  sides,  a  parallelogram  will  be  formed, 
the  perimeter  of  which  will  equal  the  sum  of  the  sides  parallel, 
to  which  lines  have  been  drawn. 

137 


138 


ELEMENTS   OF   GEOMETRY. 


6.  If  angle  bisectors  be  drawn  from  the  vertices  of  a  scalene 
triangle,  they  will  be  unequal ;  and  the  greater  one  will  be  drawn 
through  the  vertex  of  the  lesser  angle. 

6.  The  bisectors  of  the  base  angles  of  an  isosceles  triangle  are 
equal. 

7.  The  converse  of  the  last  exercise  is  true. 

8.  The  number  of  sides  of  a  polygon  may  be  found  when  the 
sum  of  its  interior  angles  is  three  times  the  sum  of  its  exterior 
angles. 

9.  Show  how  to  determine  a  line  parallel  to  the  parallels  of  a 
trapezoid  that  will  bisect  its 


10.  The  lines  joining  the 
middle  points  of  the  adjacent 
sides  of  any  quadrangle  will 
form  a  parallelogram. 

11.  The  lines  joining  the 
middle  points  of  the  adjacent 
sides  of  a  rectangle  will  be 
a  rhombus. 


Fig.  176. 


12.  The  lines  joining  the  middle  points  of  the  adjacent  sides 
of  a  rhombus  will  be  a  rectangle. 

13.  If  two  A,  A  and  B,  of  the 
A  ABC  be  bisected,  and  a  line  BE 
be  drawn  through  /,  ||  to  AB,  DE 
=  AD-\-  BE. 

14.  The  sum  of  the  perpendicu- 
lars from  any  point  in  the  base  of  an 
isosceles  triangle  to  the  equal  sides 
is  constant,  and  equals  the  altitude  from  one  of  the  equal  angles. 

Analysis.  — If  DE  +  DF=  AH,  and  if  we  should  through  D 
draw  a  parallel  to  CB,  it  would  determine  KH  =  DF,  and  in 
order  that  the  proposition  be  true,  AK  would  have  to  equal  DE. 


Fig.  177. 


PROBLEMS. 


139 


Fig.  178. 


Proof.  — AK  does  equal  DE,  because  the  ^AKD  and  DEA 
are  equal ;  being  right  triangles,  having  a  common  hypothenuse, 
and  the  Z  KDA  =  Z  EAD. 

:.  AK=DE,a.ndAK+KH=DE+DF.  q.b.  d. 

15.  A  median  of  a  triangle  is 
less  than  half  the  sum  of  the  two 
sides  which  form  the  vertex  from 
which  the  median  is  drawn. 

16.  The  angles  formed  by  the 
bisectors  of  two  angles  of  a  tri- 
angle will  be  such  that  one  of 
them  will  equal  90°  plus  half  the 
third  angle  of  the  triangle,  and  the  other  will  equal  90°  miniis 
half  the  third  angle  of  the  triangle. 

17.  In  any  right  triangle  the  median  from  the  right  angle  equals 
half  the  hypothenuse. 

18.  If  through  the  three  vertices  of  a  triangle  the  bisectors  of 
the  exterior  angles  be  drawn,  four  new  triangles  will  be  formed, 
which  will  be  mutually  equiangular. 

19.  If  the  opposite  angles  of  a  quadrangle  are  equal,  the  figure 
will  be  a  parallelogram. 

20.  A  billiard  ball  shot  parallel  to  one  of  the  diagonals  of  a 
table,  and  making  the  angles  of  reflection  equal  to  the  angles  of 
incidence,  will  return  to  the  place  from  which  it  is  shot. 


21.  If  a  diagonal  of  a  parallelogram  be  drawn,  and  lines  be 
drawn  from  the  middle  points  of  a  pair  of  opposite  sides  to  the 
other  vertices,  they  will  trisect  the  diagonal. 


140  ELEMENTS   OF   GEOMETRY. 

22.  The  bisectors  of  the  angles  of  a  quadrangle  form  a  second 
quadrangle,  the  opposite  angles  of  which  are  supplementary, 

SOME  CONSTRUCTIONS. 
(Do  not  neglect  the  analysis  or  the  discussion.) 

23.  Find  on  a  given  line  a  point  that  shall  be  equally  distant 
from  two  given  points. 

24.  Find  on  a  given  circle  points  that  shall  be  equally  distant 
from  two  given  points. 

25.  Trisect  a  right  angle. 

26.  Trisect  an  angle  of  72°. 

Note.  — The  trisection  of  any  angle  is  a  construction  that  can- 
not be  made  by  the  geometry  of  the  straight  line  and  circle,  but 
can  be  accomplished  by  the  use  of  a  number  of  curves,  the  treat- 
ment of  which  is  beyond  the  scope  of  this  book. 

27.  In  a  /SABC  draw  DE  ||  to  AC, 
so  that  DB  shall  equal  AD  +  CE. 

28.  Construct  an  equilateral  triangle, 
having  given : 

(a)  The  altitude. 

(6)  The  perimeter. 

(c)  The  radius  of  the  inscribed  circle. 

(d)  The  radius  of  the  circumscribed  p^^  jg^ 
circle. 

29.  Construct  an  isosceles  triangle,  having  given  : 
(a)  The  base  and  perimeter. 

(6)  The  base  and  altitude. 

(c)  The  base  and  angle  at  the  vertex.  ' 

(d)  The  altitude  and  the  perimeter. 

(e)  The  altitude  and  the  angle  at  the  vertex. 
80.   Construct  a  triangle,  having  given : 

(a)  Two  sides  and  their  included  median. 
(6)  An  angle,  a  side  opposite,  and  the  difference  of  the  adja- 
cent sides. 


PROBLEMS.  141 

(c)  The  base,  the  angle  at  the  vertex,  and  the  altitude. 

(d)  The  base,  the  angle  at  the  vertex,  and  the  perpendicular 
from  one  of  the  base  angles  to  the  opposite  side. 

(e)  The  base,  the  angle  at  the  vertex,  and  the  median  to  the 
base. 

(/)  The  angles  and  the  perimeter. 

(g)  Two  sides  and  the  altitude  to  the  vertex  they  form. 

{h)  Two  sides  and  the  altitude  on  one  of  them. 

(i)  An  angle,  the  side  opposite,  and  the  sum  of  the  other  two 
sides. 

( j)  An  angle  of  90°,  and  the  radii  of  the  inscribed  and  circum- 
scribed circles. 

(i-)  The  centres  of  the  escribed  circles. 

(I)  The  three  altitudes. 

31.  At  the  extremity  of  a  given  straight  line  segment  to  con- 
struct a  perpendicular  without  producing  the  segment. 

32.  Through  a  given  point  within  a  circle  to  draw  the  shortest 
chord. 

33.  Find  the  centre  of  a  circle  of  given  radius  that  shall  pass 
through  two  given  points. 

34.  In  a  given  circle  draw  a  concentric  circle,  so  that  the  diam- 
eter of  the  inner  circle  shall  equal  a  chord  of  the  outer  circle  drawn 
tangent  to  the  inner  one. 

35.  In  three  different  ways  show  how  to  construct  a  segment  of 
a  circle  that  will  contain  a  given  angle. 

36.  Let  A,  B,  and  C  represent  three  fixed  points  on  shore ;  S, 
the  point  at  which  a  sounding  _^ 

is  made  by  one  occupant  of 
a  boat,  while  another  takes 
the  angles  subtended  by  AB  -A- 

and  BC.  "  "" -^ 

Show  how  the  point  S  may 
be  located  on  a  map  which  gives 
the  positions  of  A,  B,  and  C. 

37.  Through  a  given  point 

not  in  a  given  straight  line  to  ^°"  ^^^" 

draw  a  line  that  shall  make  a  given  angle  with  the  given  line. 


142  ELEMENTS   OF   GEOMETRY. 

38.  To  a  given  circle  draw  a  tangent  that  shall  be  parallel  to  a 
given  straight  line. 

39.  In  a  given  circle  to  draw  a  chord  of  given  length  that  shall 
be  parallel  to  a  given  line. 

40.  Through  a  given  point  v?ithin  a  circle  to  draw  a  chord  of 
given  length. 

41.  Construct  a  circle  of  given  radius  that  shall  be  tangent  to 
a  given  circle,  and  also  to  a  given  straight  line. 

42.  Construct  a  circle  that  shall  cut  three  equal  chords  from 
three  given  straight  lines. 

How  will  it  be  if  these  chords  are  to  be  of  given  length  ? 

43.  Construct  the  point  through  which  the  bisector  of  the  verti- 
cal angle  of  a  triangle,  having  a  fixed  base,  will  pass. 

44.  With  the  vertices  of  a  triangle  as  centres,  construct  three 
circles  so  that  each  shall  be  tangent  to  the  other  two. 

45.  Construct  a  rectangle,  having  given  a  diagonal  and  one  side. 

46.  Will  the  sum  of  diagonal  and  radius  of  the  inscribed  circle 
be  sufficient  to  determine  a  square  ? 

47.  Construct  a  square  having  given  the  sum  of  the  radii  of  the 
circumscribed  and  inscribed  circles. 

48.  Construct  a  rectangle,  having  given  its  perimeter  and  a 
diagonal. 

49.  Construct  as  many  circles  as  possible  of  given  radius  that 
shall  be  tangent  to  a  given  circle  of  the  same  radius  and  tangent  to 
each  other. 

THEOREMS. 

60.  K  ABC  be  an  inscribed  equilateral  triangle,  and  P  be  any 
point  on  the  AB,  ^ 

PC  =  PA->rPB.  y^'^'TT^ 

61.  The  distance  of  each  side  of  the  X  /  \  ^\^ 
A^-BC(Ex.50)  from  the  centre  of  the  /  /  J^  \ 
circle  is  half  the  radius  of  the  circle.  [       /    ^^        \     i    | 

62.  If  secants  be  drawn  through  \/-^^                 \  I   / 

the    points    of    Intersection    of    two      C\ -— ____^       \l/ 

circles,  the  chords  joining  the  other  \                    ^~/Ii 

points  in  which  the  secants  intersect  ^ __^^ 

the  circles  will  be  parallel.  Fiq.  182. 


PROBLEMS. 


143 


Query.  —  How    will    it    be   if    the 
circles  are  tangent? 

53.  If  diameters  be  drawn  to  one 
of  the  points  of  intereection  of  two 
circles,  and  the  other  extremities  be 
joined  by  a  straight  line,  that  line  will 
pass  through  the  other  point  of  inter- 
section of  the  circles. 


Tig.  183. 


54.    If  PA  and  PB  be  two  tangents  to  a  circle  from  an  exterior 
point  P,  itf  be  a  moving  point  in  the  circumference,  and  QH  be  a 


Fig.  184. 


tangent  at  M,  the  perimeter  of  the  AP§B"will  be  constant  and 
=  PA-\-  PB. 

Bemark.  —  In  the  discussion  do  not  limit  Mto  the  smaller  AB. 
Look  for  positive  and  negative  distances. 

55.  The  angle  at  C  (Ex.  54)  subtended  by  QH  will  be  constant 
and  <  90°  when  M  is  on  the  smaller  AB  ;  it  will  also  be  constant, 
but  >  90°,  when  M  is  on  the  larger  AB. 

56.  If  in  a  circle  two  equal  chords  intersect,  the  segments  of 
one  will  be  equal  to  the  segments  of  the  other. 


144 


ELEMENTS   OF   GEOMETBY 


67.   If  the  angles  P  and  Q  of  the  quadrilateral  ABPCQD  be 
bisected  by  PI  and  QI, 
ZI  =  ZA  +  Za  +  Z 
also, 

^j_ZA+ZC 
2 

UZA  +  ZC=180°, 
ZI=  00°. 

Bemark. — In  the  lat- 
ter case  a  circle  could 
be  circumscribed  about 
the  quadrangle  ABCD. 


Fie.  186. 


68.  The  portions  of  any  straight  line  included  between  the  cir- 
cumferences of  concentric  circles  will  be  equal. 

69.  If  a  quadrangle  be  circumscribed  about  a  circle,  the  sum  of 
a  pair  of  non- adjacent  sides  will  equal  the  sum  of  the  other  pair. 

60.  The  converse  of  Ex.  59  is  true.  ^ 

61.  The  common  secants  of  three  intersecting  circles  are  con- 
current. 

See  Exs.  10  and  11,  §  91. 

62.  The  three  circles  which  pass 
through  the  vertices  of  a  triangle,  and 
intersect  two  and  two  on  the  sides, 
will  all  pass  through  a  common  point. 

Establish  the  converse. 


Fig.  186. 


PROBLEMS. 


145 


63.  If  BD  bisects  ZABC,  and  BE  is  a  ±  to  AC,  the 

Z  DBE  =  -^^H^. 

2 

64.  If  from  two  points,  P  and  §, 
lines  be  drawn  to  a  moving  point  M 
in  the  line  AB,  the  sum  of  PM  and  QM 
will  be  a  minimum,  when  they  make 
equal  angles  with  AB. 

Query.  —  How  will  it  be  if  the  points  are  on  opposite  sides 
of  AB  ? 

SOME  PROBLEMS  IN  LOCI. 

Note. — In  general,  to  attack  a  problem  in  loci,  make  con- 
structions enough  under  the  given  conditions  to  suggest  a  locus. 
Then  follows  the  analysis  and  the  demonstration :  If  the  suggested 
locus  be  the  required  one,  any  (every)  position  on  this  suggested 
locus  will  fulfil  the  required  conditions  ;  and  any  (every)  position 
not  on  the  suggested  locus  will  not  fulfil  the  required  conditions. 

65.  Find  the  locus  of  the  middle  point  of  a  given  straight  line 
segment,   the    extrem- 
ities of  which  remain 
in   two    lines  at  right 
angles  to  each  other. 

Let  the  lines  OY 
and  OX  be  at  right 
angles  to  each  other; 
and  let  S  be  the  given 
segment,  of  which  M  is 
the  middle  point. 

Seven  or  eight  con- 
structions under  the 
conditions  of  the  prob- 
lem are  enough  to  sug- 
gest that  the  locus  is  ^'®-  i^®- 
the  circumference  of  a  circle,  with  its  centre  at  0,  and  its  radius 
equal  to  half  the  given  segment. 


146         ELEMENTS  OF  GEOMETRY. 

If  the  suggested  locus  be  the  required  one,  any  point,  as  P,  on 
the  circumference  will  fulfil  the  required  conditions ;  and  any 
point,  as  Q,  not  on  the  circumference,  will  not  fulfil  the  required 
conditions. 

With  P  as  a  centre  and  half  of  <S^  as  a  radius  construct  an  arc 
intersecting  OX  at  H.  Join  Pand  H;  draw  the  auxiliary  segment 
OP;  and  produce  HP  until  it  intersects  Oy  at  ^ 

The  A  OPH  is  isosceles 
(two  sides  being  equal). 
The  A  OPK  is  isosceles 
(two  angles  being  equal). 
.-.  PH^PO  =  PK 
:.  HK=  S  and  is  bisected  at  P. 

To  establish  the  fact  that  Q  will  not  fulfil  the  required  con- 
ditions proceed  in  the  same  way : 

With  ^  as  a  centre  and  half  of  ^S"  as  a  radius,  construct  an  arc 
intersecting  OX  at  J.     Draw  the  auxiliary  OQ. 

QO<QJ. 
:.  ZQJO  <ZQOJ. 
.:  ZQNO>ZQOy. 
:.  QN<QO<QJ. 

That  is :  a  segment  of  length  S  cannot  be  drawn  through  Q,  so 
as  to  be  bisected  at  Q,  and  have  its  extremities  in  the  given  lines. 

The  student  will  show,  after  the  same  manner,  that  a  point 
without  the  circle  will  not  fulfil  the  required  conditions ;  and  so 
cannot  be  a  point  on  the  locus. 

Note. — Any  other  point  of  the  line  of  given  length  will  gener- 
ate a  curve,  but  that  curve  will  be  an  ellipse,  the  shape  of  which 
will  depend  upon  the  point  selected. 

66.  Find  the  locus  of  the  middle  points  of  the  chords  of  given 
length  in  a  given  circle. 


PROBLEMS. 


147 


67.  Find  the  locus  of  the  point  P,  as  M  moves  about  the  cir- 
cumference; PM  remaining  constant  in  n 
length,  and  parallel  to  its  initial  position. 

68.  Find  the  locus  of  the  middle 
points  of  all  straight  line  segments  in- 
cluded between  two  given  parallels. 

69.  Find  the  locus  of  the  middle 
points  of  all  chords  that  can  be  drawn 
through  a  given  point  in  a  circle.  Fio.  190. 

70.  Find  the  locus  of  the  middle  points  of  all  secants  through  a 
given  point  to  a  given  circle. 

71.  Find  the  locus  of  the  centre  of  a  circle  of  given  radius  that 
rolls  on  a  straight  line. 

72.  Find  the  locus  of  the  cen- 
tre of  a  circle  of  given  radius 
that  rolls  on  the  circumference 
of  a  given  circle. 

73.  AB  is  a  diameter  of  a  cir- 
cle ;  through  A  a  secant  is  drawn  ; 
at  T,  a  tangent,  and  through  B  a 
secant  perpendicular  to  the  tan- 
gent. Find  the  locus  of  the  in- 
tersection of  the  two  secants. 

74.  Let  e  change  from  0^=  to  360°,  and  CP 
locus  of  P. 


Find  the 


Fig.  192. 


Fro.  193. 


75.    AB  is  a  fixed  line  ;  and  F  a  given  point. 

PP  JVl 

Locate  P,  so  that  —  =  —   (constant). 
PH      n  ^ 


148 


ELEMENTS   OF   GEOMETRY. 


76.    AB    and    CD    are 

given  lines.  Find  the  locus 
of  P,  which  moves  so  that 
the  sum  of  the  perpendic- 
ulars on  the  given  lines 
equals  q. 

Note.  —  When  perpen- 
diculars to  a  given  line 
are  being  considered,  the 

distances  on  one  side  are  positive,  and  on  the  other  side  negative. 
This  should  be  borne  in  mind  in  74  and  76. 


iP 

I 


K      B 


77.  AB  and  CD  are  fixed  at  right 
angles  to  each  other.  Find  the  locus 
of  P,  PZf "'  +  PK'  being  constant. 

78.  A  triangle  has  a  given  base 
and  vertical  angle.    Find  the  locus     A 
of  its  vertex. 

79.  Find  the  locus  of  the  centre 
of  the  inscribed  circle  ;  the  triangle 
being  as  described  in  Ex.  78. 

80.  In  the  same  triangle,  find  the  locus  of  the  point  of  inter- 
section of  the  altitudes. 

81.  Find  the  locus  of  the  centre 
of  a  circle,  tangent  to  two  given 
straight  lines. 

Hemark.  —  Do  not  fail  in  each 

problem  to  discuss  the  particular    

cases.  -^ 

82.  Find  the  locus  of  a  point, 
the  sum  of  the  distances  of  which 
from  two  vertices  of  an  equilateral 
triangle  equals  its  distance  from  the 
third  vertex. 


C 

Fig.  195. 


C 

M 


N 


B 


D 

Fig.  196. 


83.    Find  the  locus  of  points  from  which  a  given  circle  will  sulv 
tend  a  given  angle. 


PROBLEMS. 


149 


84.  MNP  is  a  right  triangle,  the  oblique  vertices  of  which 
remain  in  AB  and  CD,  that  are  fixed  at  right  angles  to  each  other. 
Find  the  locus  of  P. 

85.  TJ^  +  Tb^z=AB^.    Find  the  locus  of  p. 


Fig.  197. 


86.   ABC  is  a  given  right  triangle  ;   HN  is  a  moving  perpen- 
dicular ;  AM  and  CN  meet  at  P.     Find  the  locus  of  P. 


SOME  PROBLEMS   IN  CONSTRUCTION. 

87.  Through  a  given  point  within  an  angle  to  draw  a  line  so 
that  the  segments  included  between  the  lines  shall  be  bisected  at 
the  point. 

Let  IR  and  IQ  represent  the 
lines  forming  the  angle,  and  G 
the  given  point. 

Analysis.  —  If  HK  were  the 
required  line  (without  saying 
whether  it  is  or  is  not),  and  if 
through  G  an  auxiliary  line  were 
drawn  ||  to  ///,  it  would  bisect 
IK ;  or  if  the  auxiliary  line  were 
drawn  ||  to  IK,  it  would  bisect   IR.     Hence  the 

Constniction.  —  Through  the  given  point  draw  an  auxiliary  line 
parallel  to  one  of  the  lines  forming  the  angle.  Let  its  intersec- 
tion with  the  other  one  be  A.  Lay  off  AH=  lA,  and  draw  the 
line  HG.     The  segment  IIK  will  be  bisected  at  G  as  required. 


Fig.  199. 


150 


ELEMENTS   OF   GEOMETRY. 


Fig.  200. 


Discussion.  —  It  G  should  lie  in  an  angle  bisector,  the  A  KIH 
would  be  isosceles.  If  G  should  lie  in  one  of  the  lines  forming 
the  angle,  HK  would  coincide  with  that 
line.  If  G  should  coincide  with  /,  there 
would  not  be  any  construction. 

Note.  —  Exercise  87  has  thus  been  given 
in  detail,  not  because  of  its  difficulty,  but 
because  it  is  desirable  to  present  to  the 
student  the  method  of  investigation  that  is 
to  be  pursued  in  all  problems  of  construc- 
tion, whether  they  be  easy  or  difficult. 

While  in  general,  in  the  text-book  solu- 
tion of  a  problem,  but  one  figure  is  used 
for  the  Analysis  and  the  Proof,  it  is  advisable  that  the  student 
should  use  two. 

88.  AB  and  CD  are  fixed  at  right  angles  to  each  other,  and 
EF  is  any  other  line.     To  con- 
struct a  square  such  that  two 
sides  shall  lie  on  AB  and  CD, 
and  one  vertex  shall  lie  on  EF. 

89.  Construct  a  square,  hav- 
ing given : 

(a)  The  diagonal. 

(b)  The  sum  of  a  diagonal 
and  one  side. 

(c)  The  difference  of  a  di- 
agonal and  one  side. 

90.  Construct  a  rectangle,  having  given : 

(o)  A  diagonal  and  the  difference  of  two  adjacent  sides. 
(6)  One  side  and   the  corresponding   angle   formed    by  the 
diagonals. 

(c)  One  side  and  the  sum  of  the  diagonals. 

91.  Construct  a  parallelogram,  having  given : 
(a)  One  side  and  the  diagonals. 

(6)  The  diagonals  and  one  angle  of  the  parallelogram, 
(c)  One  side,  one  diagonal,  and  one  angle. 


PROBLEMS. 


161 


92.  Construct  a  circle  of  given  radius,  that  shall  intersect  a 
given  circle,  so  that  they  shall  be  at  right  angles  to  each  other  at 
a  given  point. 

93.  Construct  a  circle  with  a  given  centre  which  shall  intersect 
a  given  circle  at  right  angles. 

94.  Construct  a  circle  of  given  radius  : 

(a)  That  shall  be  tangent  to  two  given  straight  lines. 

(6)  That  shall  be  tangent  to  a  given  line  and  to  a  given  circle. 

(c)  That  shall  be  tangent  to  two  given  circles. 

(d)  That  shall  pass  through  a  given  point,  and  be  tangent  to  a 
given  line. 

(e)  That  shall  pass  through  a  given  point,  and  be  tangent  to  a 
given  circle. 

95.  Construct  a  circle  which  shall  be  tangent : 

(a)  To  two  given  lines,  and  pass  through  a  given  point. 
(6)  To  a  given  line  at  a  given  point,  and  that  shall  pass  through 
another  given  point. 

(c)  To  a  given  circle  at  a  given  point,  and  tangent  to  a  given 
straight  line. 

(d)  To  a  given  circle  at  a  given  point,  and  shall  pass  through 
another  given  point. 

(e)  To  a  given  line  at  a  given  point,  and  also  tangent  to  a 
given  circle. 

Let  C  represent  the  centre 
of  the  given  circle,  AG  the 
given  line,  and  G  the  point  at 
which  the  required  circle  is  to 
be  tangent. 

Analysis.  —  ij^'the  circle,  the 
centre  of  which  is  H,  xoere  the 
required  circle,  and  j/"  at  G*  a 
perpendicular  loere  erected  it 
would  pass  through  H.  CT 
would  also  pass  through  H, 
and  TG  would  form  with  TH 
and  GH  an  isosceles  triangle. 


Fig.  202. 


Then  i/ through  C  a  parallel  to  TG 


152  ELEMENTS   OF   GEOMETRY. 

were  drawn,  it  would  form  an  isosceles  ACHK,  and  GK  would 
equal  CT. 

Construction.  — At  the  given  point  of  the  given  line  erect  a  per- 
pendicular ;  on  the  opposite  side  of  the  line  from  the  given  circle, 
lay  off  GK,  the  radius  of  the  given  circle.  Draw  CK.  At  C  con- 
struct an  Z  KCH  =  Z  CKG.  It  will  determine  H,  the  centre  of 
the  required  circle. 

With  Has  a,  centre  and  a  radius  HG,  construct  a  circle  ;  it  will 
fulfil  the  required  conditions.  (H  might  have  been  as  well  deter- 
mined by  a  perpendicular  erected  at  the  middle  point  of  CK;  or  T 
might  have  been  determined  by  drawing  through  G  a  parallel 
to  KC.) 

Discussion.  —  If  a  circle  tangent  to  AG  at  G  should  increase  in 
radius,  beginning  with  a  radius  (0),  it  would,  when  the  i-adius 
equals  HG,  be  tangent  to  the  given  circle.  Beyond  that  it  would 
be  an  intersecting  circle  for  a  time,  and  then  in  one  position  be 
tangent  to  the  given  circle.  Beyond  that  it  would  be  an  envelop- 
ing circle. 

Let  us  now  see  about  the  second  solution  that  has  been  sug- 
gested by  the  discussion. 

Analysis.  —  If  the  circle  the 
centre  of  which  is  H'  were  an 
enveloping  tangent  circle,  and 
also  a  tangent  to  .46^  at  G' ;  a 
I)erpendicular  erected  at  G  would 
pass  through  H' ;  T'C  would 
pass  through  H' ;  the  A  T'H'G 
would  be  isosceles  ;  and  CK'  \\  to 
2^6?  would  determine  GK'  =  CV. 
Hence  the 

Constrriction.  —  Erect  a  perpendicular  at  G.  Lay  off  GK',  on 
the  side  toicard  the  given  circle,  equal  to  the  radius  of  the  given 
circle.  Draw  CK'.  Through  G  draw  a  parallel  to  CK',  thus  deter- 
mining T'.  The  intersection  of  T'C  with  the  perpendicular  at  G, 
determines  H',  and  H'G  is  the  radius  of  the  required  circle. 

With  //'  as  a  centre,  and  a  radius  H'G  construct  the  required 
circle. 


PROBLEMS. 


153 


Further  Discussion.  —  If  AQ-  be  tangent  to  the  given  circle 
at  §,  there  will  be  but  one  construction. 


Fig.  204. 


If  G  and  Q  coincide,  there  will  be  an  infinite  number  of 
solutions. 

If  ^G^  intersects  the  given  circle,  and  &  is  without,  there  will 
be  two  constructions. 


Fig.  206. 

li  AG  intersects  the  given  circle,  and  G  is  on  the  circumference, 
there  will  not  be  any  solution. 

If  G  should  lie  within  the  given  circle,  there  will  be  two  con- 
structions. 

In  order  to  reach  these  different  constructions,  AG  may  be  con- 
ceived as  moving  from  one  position  to  another  that  is  parallel ;  and 
the  point  G  may  be  conceived  as  moving  in  the  line  AG.  It  is 
interesting  to  note  the  special  cases ;  i.e.  the  point  and  the  straight 


154  ELEMENTS    OF   GEOMETRY. 

line  toward  which  the  circles  approach  as  G  approaches  some  of 
its  particular  positions. 


Fig.  20T.  Fio.  208. 

96.  Through  a  point  to  draw  a  line  that  shall  meet  two  given 
lines,  which  intersect  outside  the  limits  of  the  drawing.  In  practice 
on  land  it  would  read :  "  Which  intersect  in  some  inaccessible 
point." 

97.  Construct  a  circle  that  shall  be  tangent  to  a  given  line  at  a 
given  point,  and  shall  pass  through  a  given  point. 

98.  Pass  the  circumference  of  a  circle  through  two  points,  and 
tangent  to  a  given  straight  line. 

99.  Construct  a  triangle,  having  given  one  angle,  the  radius  of 
the  inscribed  circle,  and  the  radius  of  the  corresponding  escribed 
circle. 

100.  Construct  a  triangle,  having  given  one  vertex,  and  the  feet 
of  two  altitudes. 

101.  Construct  a  triangle,  having  given  the  feet  of  the  altitudes. 

102.  Use  some  of  the  following  figures  to  establish  the  fact  that 
the  square  on  the  hypothenuse  of  a  right  triangle  equals  the  sum  of 
the  squares  on  the  other  sides. 

Note.  —  These  figures  have  come  from  various  sources,  princi- 
pally, however,  from  students  who  have  been  called  upon  for 
original  demonstrations.  The  number  might  have  been  largely 
increased,  but  a  large  enough  number  lias  been  presented  to  show 
that  there  may  be  more  than  one  way  of  solving  a  problem. 


PBOBLEMS. 


155 


Fig.  209. 


156 


ELEMENTS  OF  GEOMETRT. 


PROBLEMS. 


167 


^  =  ^     or  BC^  =  AC-DC 
DC     BC 

4^  =  4?     or  AB^  =  AG  AD 
AD     AB 


BC'  +  AB-  =  AC  (AD  +  DC) 

Fig.  211 


\ 

^\ 

\ 

\ 

^^^ 

C 

A 

B 

(10) 

168 


ELEMENTS   OF   GEOMETRY. 


Fio.  212. 


PROBLEMS. 


169 


Fio.  213. 


160 


ELEMENTS   OF   GEOJVIETRY. 


Fig.  214. 


PROBLEMS. 


161 


Fie.  215, 


162 


EliEMENTS   OF   GEOMETRY. 


a'- 

^^"^ 

\ 
\ 
\ 
\ 
\ 

V 

\ 

\ 

> 

b' 

Fio.  216. 


The  most  noted  figures  are:    (1),  (3),  (5),  (6),  (25),  and  (27). 
The  simplest  demonstrations  are  made  by  the  figure  already 
used  in  the  body  of  the  text,  and  by  (6),  (G),  (24),  and  (39). 


PROBLEMS.  163 


PROBLEMS. 

103.  How  many  trees  can  be  planted  on  160  acres  of  land : 

(ffl)  When  set  out  in  equilateral  triangles,  each  side  being  20 
feet? 

Note. — This  manner  of  placing  trees    •       •       •       •       • 
is    sometimes    called    the    hexagonal 

method  ;  as  any  given  tree  (not  on  the         •       •       •        •        • 
outer  boundary)  will  have  six  trees  at 

the  same  distance  from  it,  forming  the    •       •       •       •       o 
vertices  of  a  hexagon,   of  which  the 
given  tree  is  the  centre.  •       •        •       • 

(6)  When  set  out  in  squares,  each  side  being  20  feet  ? 

(c)  When  planted  quincunx,  20  feet  apart  ?  * 

104.  What  regular  polygons  may  be  used  by  themselves  to 
cover  a  plane  surface  ? 

What  combinations  of  regular  polygons  may  be  used  for  the 
same  purpose  ? 

105.  Why  does  the  honey  bee  build  a  hexagonal  cell  ? 

C 


106.    To  convert  a  polygon  into  an  equivalent  triangle. 

(a)  If  the  given  polygon  be  convex,  and  it  be  desired  to  have 
one  side  of  the  triangle  in  the  line  of  one  side  of  the  polygon, 
as  AF: 

*NoTE.  —  Any  tree  (not  on  the  outer  boundary)  will  be  the 
centre  of  a  square,  each  vertex  of  which  will  be  20  feet  from  the 
centre. 


164 


ELEMENTS   OF   GEOMETRY. 


Cut  off  the  A  ABC,  and  put  on  its  equivalent  A  AGO. 
Cut  off  the  A  OCD,  and  put  on  its  equivalent  A  GHD. 
Cut  off  the  A  FED,  and  put  on  its  equivalent  A  FKD. 
The  polygon  is  reduced. 

(6)  If  the  given  polygon  be  re-entrant,  put  on  and  cut  off,  until 
it  is  changed  to  a  convex  polygon,  and  then  proceed  as  before. 

107.  Show  that  the  diagonals  of  a  trapezoid  separate  it  into 
four  triangles,  two  of  which  are  similar,  and  the  other  two  are 
equivalent  in  area. 

108.  Depending  on  the  properties  of  similar  triangles,  show  how 
to  determine  the  direction  in  which  to  run  a  tunnel  so  that,  start- 
ing at  a  given  point  it  shall  run  in  a  straight  line  under  a  mountain, 
and  shall  emerge  from  the  surface  at 

another  given  point. 


109.  Show  geometrically  how  to  find 
the  distance  across  a  river  in  6  different 
ways. 

Do  not  neglect  the  simplest  one. 


Fig.  218. 


110.   Construct  a  scale  that  shall  measure  hundredtha 


1  2 

Fig.  219. 


111.    Construct  a  vernier  that  shall  measure  hundredths. 

Let  the  space  from  ^  to  i?  represent  one  of  the  divisions  of  a 
scale  which  is  subdivided  as  indicated  in  the  drawing  into  10 
equal  parts  (if  we  use  the  decimal  subdivision). 


PROBLEMS. 


165 


The  vernier  slides  along  on  the  scale  and  has  the  distance  PQ 
(which  in  length  equals  9  of  the  subdivisions  on  the  scale) ,  sepa- 
rated into  10  equal  parts.  Then  each  subdivision  on  the  vernier 
will  be  less  by  one-tenth  than  each  subdivision  on  the  scale. 

Let  us  read  the  distance  of  the  point  P, 
from  the  0  (zero)  end  of  the  scale.  It  is 
3.18,  an  approximation,  of  course,  but  the 
error  is  less  than  .01.  The  student  will 
supply  the  reasons  for  the  reading. 

Note.  —  Verniers  are  in  great  variety, 
and  are  called  into  use  in  cases  where 
accuracy  of  observation  is  required.  The 
levelling  rods  of  the  surveyor  read  to  thou- 
sandths of  a  foot,  and  some  of  the  circles 
used  by  surveyors  in  the  measurement  of 
angles  read  to  ten  seconds. 


bj- 


^« 


o 

> 

r- 
m 


So 


oV^ 


m 

z 
m 
30 


112.  If  a  piece  of  paper  8  units  square  be  cut  as  indicated  in 
(a),  and  then  be  placed  as  in  (&),  64  square  units  apparently 
become  Go  square  imits.     Where  is  the  deception? 


166 


ELEMENTS   OF   GEOMETRY. 


113.  Construct  a  square  that  shall  have  one  side  on  the  base  of 
a  triangle,  and  its  other  vertices  in  the  other  two  sides  of  the 
triangle. 


Fig.  222. 

114.   If  straight  lines  be  drawn  from  the  vertices  through  any 
point  within  a  triangle, 

PB     QC     KA 

116.  If  from  any  point 
within  a  triangle,  straight 
lines  be  drawn  to  the 
sides,  and  through  the 
vertices  lines  be  drawn 
parallel  to  them,  then  will 

DP  ^  EP  ^   FP^i 

HB  KG  MA 
116.  (a)  Show  that  a 
point  may  be  located  be- 
tween the  extremities  of 
a  given  segment,  so  that 
the  ratio  of  its  distances 
from  the  extremities 
shall  equal  any  assumed 

ratio.  Fig.  224. 

(6)  Show  that  a  point  on  the  line,  but  exterior  to  the  segment, 
may  be  found  which  will  fulfil  the  same  conditions. 


PROBLEMS. 


167 


117.  Approximately,  what  will  be  the  relative  amounts  of  water, 
that  under  the  same  head  will  be  discharged  from  an  inch  pipe  and 
from  a  quarter  inch  pipe. 

118.  Having  given  one  side  of  a  decagon  10,  find  the  radius. 

119.  One  side  of  a  pentagon  is  10  ;  find  a  diagonal. 

120.  The  radius  of  a  circle  being  1,  find  the  area  of  an  inscribed 
regular : 


(o)  Triangle. 
(6)  Quadrangle. 

(c)  Pentagon. 

(d)  Hexagon. 


(e)  Octagon. 
(/)  Decagon. 
(g)  Dodecagon. 


121.  The  radius  of  the  circumscribed  circle  being  10,  find  the 
side  and  the  apothem  of  the  regular  inscribed  decagon. 

122.  Compare  the  areas  of  the  following  regular  figures,  which 
have  the  same  perimeter : 


(a)  Triangle. 
(6)  Quadrangle. 

(c)  Pentagon. 

(d)  Hexagon. 


(e)  Octagon. 
(/)  Decagon, 
(jf)  Dodecagon. 
(A)  Circle. 


123.  Three  circumferences,  each  of  radius  r,  are  tangent  to  each 
other.     Find  the  included  area. 

124.  The  front  and  rear  wheels  of  a  carriage  are  respectively  3 
and  4  feet  in  diameter.     Will  the  two 

points  that  are  uppermost  at  the  start 
again  be  uppermost  at  the  same  time,  if 
the  carriage  should  travel  on  a  straight 
line? 

125.  Show  that  in  the  accompanying 
figure,  d  will  be  a  side  of  an  inscribed 
decagon,  and  p  will  be  a  side  of  an  in- 
scribed pentagon. 


168 


ELEMENTS  OP  GEOMETRY. 


THEOREMS. 

126.  From  any  point  within  a  regular  polygon,  if  perpendicu- 
lars be  let  fall  on  the  lines  forming  the  sides,  their  average  length 
will  be  that  of  the  apothem  of  the 
polygon. 

127.  If  on  each  of  the  three  sides  of  a 
right  triangle,  as  diameters,  semicircles 
be  constructed,  as  indicated  in  the  figure, 
the  sum  of  the  areas  of  the  two  crescents 
will  equal  the  area  of  the  right  triangle.  fig.  226. 

128.  If  four  circles  are  constructed  as  in  the  figure,  the  radius 
of  the  small  circles  wiU  be  one-third  of  the  radius  of  the  large  circle. 


Fio.  22T. 


Fig.  228. 


129.  If  the  diameter  of  a  circle  be  separated  into  any  two  parts 
and  semicircles  be  constructed  as  indicated, 

(a)  AB  +  BC  =  APC. 

(b)  The  areas  into  which  the  circle 
is  separated  by  the  arcs  AB  and  BG 
will  be  to  each  other  as  AB  is  to  BC 

130.  The  area  of  the  ring  bounded 
by  two  concentric  circumferences 
equals  the  area  of  a  circle,  the  diame- 
ter of  which  is  a  chord  of  the  larger 
that  is  tangent  to  the  smaller.  Fio.  229. 


PROBLEMS. 


169 


131.  If  CA  be  separated  into  any  number  of  equal  parts,  per- 
pendiculars at  the  points  of  division  be  drawn  to  intersect  the  cir- 
cumference constructed  on  CA 
as  a  diameter,  and  these  points 
of  intersection  be  joined  with 
C,  they  will  be  radii  that  will 
determine  rings  of  equal  area. 

132.  If  the  centres  A  and 
B  of  two  circles  be  joined,  and 
a  circle  of  any  radius  a,  with 
the  middle  of  AB  as  its  centre, 
be  constructed,  the  sum  of  the 
squares  of  the  tangents  to  the 
first  circles  from  any  point  in 
the  circumference  of  the  third 
circle,  will  be  constant. 

133.  If  G  represent  the 
point  of  intersection  of  the 
medians  of  a  A  ABC,  and 
P  be  any  point  in  the  plane, 

PA^  +  PB"  +  PO^ 

=  AG^  +  BG^+  Ca^ 

+  .3  PG"^- 

Note.  —  (r  is  the  centre  of 
gravity  of  the  triangle. 

134.  If  any  number  of  par- 
allels intersect  two  straight  lines,  the  middle  points  of  these 
parallel  segments,  together  with  the  points  of  intersection  of  the 
diagonals  of  the  trapezoids  formed,  all  lie  in  the  same  straight  line. 

135.  In  any  triangle,  the  middle  of  each  side,  the  feet  of  the 
three  altitudes,  and  the  middle  points  of  the  segments  of  the  alti- 
tudes between  their  common  intersection  and  the  vertices  of  the 
triangle,  are  9  points  in  the  circumference  of  a  circle. 


Fig.  281. 


Note.  —  This  is  one  of  the  famous  theorems. 


170 


ELEMENTS   OF   GEOMETRY. 


Suggestion.  —  A  simple  method  of  establishing  the  theorem  is 
by  the  use  of  quadrangles. 

Show  that  the  circle  which 
passes  through 

(a),  1,  2,  3,  will  pass  through  4. 
(6),  2,  3,  4,  will  pass  through  5, 
(c),  2,  3,  5,  will  pass  through  6. 
(d),  1,  2,  3,  will  pass  through  7. 
(e),  2,  5,  6,  will  pass  through  8. 
(/),  2,  3,  4,  will  pass  thi-ough  J>, 

Find  the  centre. 

PROBLEMS  IN  LOCI. 

136.   Find  the  locus  of  a  point  which  moves  so  that  the  ratio  of 
its  distances  from  two  fixed  points  always  remains  the  same. 

Let  A  and  B  represent  the  two  fixed  points.    By  Ex.  116,  two 

positions  will  lie  on  AB,  jn 

one  between  A  and  B^ 
and  the  other  outside 
AB. 

m 


Fig.  232. 


Let  —   represent 
n 

fixed    ratio 

the    points 

will  lie  as 

D  and  C. 


the 

If  wi  >  n, 
on  the  line 
indicated  at 


Fio.  238. 

M^JH^ndiGA^m 
DB     n         CB     n 

A  number  of  constructions  made  under  the  given  conditions 
suggest  that  the  locus  is  the  circumference  of  a  circle  on  Z>C  as  a 
diameter. 

Analysis. — i/ the  suggested  figure  be  the  required  locus,  any 
point  on  the  circumference  must  fulfil  the  required  conditions  and 
any  point  not  on  the  suggested  locus  must  fail  to  fulfil  the  required 
conditions  ;  i.  e.  if  P  be  any  point  on  the  circumference,  we  must 
show  that  PA  -j-  FB  —m-i-n;  and  that  if  §  be  any  point  not  on 


T^M 


PROBLEMS. 


171 


the  circumference,  we  must  show  that  QA  -i-  QB  does  not  equal 
m^n,  (QA  -f-  QB^m  -7-  n). 

Furthermore,  if  the  suggested  locus  be  the  locus,  any  line 
through  D  (other  than  a  tangent  to  the  suggested  circle)  will  have 
J)  and  another  point,  from  which  segments  of  lines  drawn  to  A 
and  to  B  will  have  the  ratio  of  m  -^  n. 

Demonstration.  —  Draw  any  line  through  2>,  as  DH. 

Draw  BF±  to  DU. 

Join  A  and  F. 

DA_m 

DB~  n 

FA^m 

FB     n 

{FA  being  >  DA,  and  FB  being  <  DB) 

By  hypothesis  —  >  1. 

71 

If  a  point  should  move  from  D  toward  F,  and  beyond  to  infinity, 
the  ratio  of  its  distances  from  A  and  JB,  after  passing  F,  would 
approach  1,  as  the  limit  of  the  ratio.  Then  somewhere  between 
F  and  00  the  ratio  would 
be  m  -^  n. 

Let  H  represent  the 
point. 

Draw  IIA  and  LIB. 

IheZAHB  is  bisected 
hy  HD. 

A  perpendicular  to  HD 
at  H  will  pass  through  C. 

If  there  be  any  point 

Q,  other  than  D  and  H 

on  the  line  DH  such  that  Fig.  284. 

QA_m 

QB~  7j' 

a  perpendicular  to  HD  erected  at  Q  would  pass  through  C,  which 

is  impossible. 

Hence  H  will  be  on  the  locus  of  the  vertex  of  a  right  triangle 

having  DC  for  its  hypothenuse.     That  locus  is  the  circumference 


172         ELEMENTS  OP  GEOMETRY. 

of  a  circle  having  DC  for  its  diameter;  any  point  Pon  the  circle 
in  Fig.  1  will  be  the  pomt  H  of  the  line  joining  P  and  D  ;  and  the 
suggested  circumference  is  the  required  locus. 

Discussion.  —  If  the  ratio  m  -r-  n  should  be  made  to  approach  1, 
the  diameter  of  the  circle,  which  is  the  locus,  would  increase, 
D  would  approach  the  middle  of  the  segment  AB,  and  C  would 
approach  oo  .  If  the  ratio  should  reach  1,  D  would  be  at  the  middle 
of  AB,  C  would  be  at  00 ,  and  what  was  a  circle  would  reach  the 
limit  toward  which  it  approached  ;  viz.  a  perpendicular  bisector  of 
the  segment  AB.  If  the  ratio  fall  below  1,  the  locus  would  be  a 
circumference  having  its  centre  on  the  opposite  side  of  A. 

The  centre  of  the  locus  depending  on  the  ratio  m  ^n  may  occupy 
any  position  on  the  line  AB  except  between  the  points  A  and  B. 

The  ratio  m  ■-  n  may  be  changed  by  causing  either  the  numer- 
ator or  the  denominator  to  change,  or  both. 

If  n  should  remain  fixed,  and  m  should  increase,  both  D  and  C 
would  approach  B ;  and  if  m  should  become  oo  ,  the  points  D  and 
C  would  coincide  with  B,  and  the  circle  would  be  reduced  to  a 
point  If  m  should  remain  fixed,  and  n  should  approach  0,  the 
same  thmg  would  result. 

If^<l, 
n 

and  m  should  decrease  while  n  remained  fixed,  the  circle  which 
now  would  encompass  A  would  decrease  in  diameter,  and  the 
circle  would  degenerate  to  a  point  at  A  when  m  should  become  0. 
If  m  should  remain  fixed  and  n  should  approach  oo  ,  the  same  thing 
would  result. 

Note.  — This  is  the  famous  "Problem  of  the  Lights."  In  this 
solution  the  case  has  been  that  of  the  locus  of  the  points  of  equal 
illumination  on  a  plane  which  passes  through  the  line  of  two  lumi- 
nous points. 

In  the  Solid  Geometry  it  will  be  shown  that  the  complete  locus 
is  the  surface  of  a  sphere  enveloping  the  lesser  light.  The  diameter 
of  the  sphere  is  determined  by  the  distance  apart  of  the  luminous 
points  and  by  the  ratio  m  h-  n. 

This  problem  is  very  prettily  handled  by  the  methods  of  the 
Analytic  Geometry. 


Fig.  285. 


PROBLEMS.  173 

137.  Given  two  circles.  Find  the  locus  of  a  point  which  moves 
so  that  the  difference  of  the  squares  on  the  tangents  from  it  to  the 
two  circles  shall  be  constant. 

Note.  —  When  the  difference  is  0,  the  locus  is  called  the  radical 
axis. 

138.  Find  the  locus  of  a  point  from  any  position  of  which  two 
given  circles  subtend  the  same 
angle. 

139.  Two   circumferences    in-   A 
tersect  at  H  and  K.     Find  the 
locus  of  the  middle  point  of  AB 
through  H, 

140.  A  circle,  the  diameter  of 
which    equals    the    radius   of    a 
larger  circle,  is  internally  tangent  to  the  larger  one  and  rolls  on  its 
circumference. 

Find  the  locus  of  any  given  point  P  on  the  circumference  of 
the  smaller  circle. 

Query.  —  If  segments  of  the  rolling 
circumference  be  of  different  colors, 
what  will  be  the  effect  of  a  very  rapid 
rolling  of  the  smaller  circle  on  the  larger 
one? 

Note.  —  This  is  a  special  case  of  a 

general  problem ;  viz.     Find  the  locus 

of  any  point  in  the  plane  of  a  circle,  as 

the  latter  rolls  on  the  circumference  of  „ 

Fig.  236. 
another  circle. 

The  general  case  is  better  treated  by  the  methods  of  Algebraic 

Geometry. 

141.  Find  the  locus  of  a  point,  the  distances  of  which  from  two 
given  straight  lines  have  a  fixed  ratio. 

142.  Find  the  locus  of  a  point  which  moves  so  that  the  tangents 
from  it  to  two  given  circles  are  equal. 


174  ELEMENTS   OF   GEOMETRY. 

148.  Find  the  locus  of  a  point  which  moves  so  that  the  sum  of 
its  distances  from  two  vertices  of  an  equilateral  triangle  shall 
equal  its  distance  from  the  third. 

144.  Find  the  locus  of  a  point  which  moves  so  that  the  svun  of 
the  squares  of  the  distances  from  two  given  points  is  fixed.  The 
same  for  three  given  points. 

146.  Find  the  locus  of  a  point  which  moves  so  that  the  differ- 
ence of  the  squares  of  its  distances  from  two  fixed  points  is  con- 
stant. 

146.  Within  a  circle  two  chords  at  right  angles  to  each  other 
intersect  in  a  fixed  point.  Find  the  locus  of  the  middle  points  of 
the  sides  of  the  quadrangle  of  which  the  perpendicular  chords  are 
the  diagonals. 


SOLID   AND   SPHERICAL   GEOMETRY. 


a>»<c 


CHAPTER   IX. 


The  first  eight  chapters  of  this  work  have  dealt  chiefly  with  the 
relations  of  figures  in  a  single  plane.  The  remainder,  making  use 
of  what  has  been  developed  in  the  Plane  Geometry,  will  not  be 
confined  to  a  plane. 

105.  Theorem.  Tlie  intersection  of  two  planes  is  a 
straight  line. 

1 

\b 


I 


E 

Fig.  28T. 


A  plane  has  been  defined  as  a  surface  such  that  if  any 
two  points  in  it  be  joined  by  a  straight  line,  the  line  will 
be  wholly  in  the  surface. 

If  two  planes  intersect,  they  will  have  more  than  one 
point  in  common.  Any  two  of  these  points  determine  a 
straight  line,  which  straight  line  will  lie  in  both  planes, 
hence  must  be  common  to  both  and  so  be  their  intersec- 
tion. Q.E.  D. 

Planes  are  usually  represented  by  a  quadrangle  which 
is  a  part  of  the  surface,  and  the  quadrangle  is  usually 

176 


176         ELEMENTS  OF  GEOMETRY. 

taken  in  the  form  of  a  parallelogram,  and  is  designated 
as  the  plane  CE,  for  instance. 

Exercises.  —  1.  Show  that  a  straight  luie  and  a  point  deter- 
mine a  plane. 

2.  Show  that  three  i)oints  determine  a  plane. 

3.  Show  that  a  pair  of  intersecting  straight  lines  determine  a 
plane. 

106.  When  two  planes  intersect,  they  form  four  diedral 
angles. 

A  diedral  angle  between  two  planes  is  the  amount  of 
rotation  that  one  plane  would  have  to  undergo  about  the 
line  of  intersection  as  an  axis,  to  coincide  with  the  other. 

The  measure  of  a  diedral  angle  is  the  plane  angle 
formed  by  two  lines,  one  in  each  plane,  perpendicular  to 
the  line  of  intersection  at  the  same  point. 

Let  MN  and  QS  represent  two  planes,  IJ  the  line  of 
intersection,  B  any  point 

in  that  line,  BA  a  line  in  ^^^      \j 

the  plane  MN  ±  to  IJ, 
and  BC  a  line  in  the 
plane  QS  A.U>  IJ. 

The    plane  Z  ABC  is 
the  measure  of  the  diedral 

M-IJ-S.  F'«-  28S. 

The  vertical  plane  Z  HBK  is  the  measure  of  the 
diedral,  vertical  to  the  one  measured  by  ABC. 

The  plane  angle  ABH,  the  supplement  of  ABC,  is  the 
measure  of  the  diedral  M-IJ-Q. 

CBK,  the  vertical  of  ABH,  is  the  measure  of  S-IJ-X, 
the  vertical  of  M-IJ-Q.     Hence  the 

Theorem.     Vertical  diedrcUs  are  equal. 


INTERSECTIONS   OF    PLANES. 


177 


107.   Diedrals  are  acute,  right,  or  obtuse  according  as 
the  measuring  plane  angle  is  acute,  right,  or  obtuse. 


Fio.  289. 

If  a  diedral  be  90°,  the  line  BA  will  be  perpendicular 
to  both  BI  and  BC,  which  are  lines  on  the  plane  QS. 

Theorem.  If  a  line  be  j^^rpendicular  to  two  lines  of 
a  plane  at  their  intersection,  it  will  he  perpendicular  to  any 
line  of  the  plane  passing  through  its  foot. 

Let  BA  be  perpendicular  to  BC  and  BH.     Let  BP  be 
any  line  of   the  plane  QS  passing 
through  B. 

Analysis.  —  If  BP  is  perpendicu- 
lar to  AK  and  only  under  that 
condition,  we  know  that  oblique 
lines  dra^ai  from  any  point  of  BP 
to  points  equally  distant  from  the 
foot  of  the  perpendicular  will  be  equal. 

Demonstration.  —  Draw  HC,  a  straight  line  in  the 
plane  QS  intersecting  the  three  lines  of  the  plane  at  H, 
P,  and  C.  Lay  off  BK=  BA  and  join  the  points  H,  P, 
and  C  with  A  and  K. 


Fig.  240. 


178  ELEMENTS   OF   GEOMETRY. 

HA  =  HK    (A  IIBA  =  A IIBK), 
CA  =  CK     (A  CBA  =  A  CBK), 

HC=nc. 

.\  AAHC=AKHC, 
and  ZACH=ZKCH 

Then  A  ACP  =  A  KCP 

(having  two  sides  and  included  angle  in  each  equal). 
•  •-  PA  =  PK, 
and  that  which  was  necessary  and  sufficient  to  establish 
the  theorem  is  proved.  q.  e.  d. 

Exercises.  —  1.  Show  that  if  a  line  perpendicular  to  another 
line  which  it  intersects,  be  rotated  about  the  second  line  as  an  axis, 
it  will  generate  a  plane  perpendicular  to  the  axis. 

2.  Find  the  locus  of  a  point  the  ratio  of  the  distances  of  which 
from  two  fixed  points  equals  one. 

108.  Definition.  A  straight  line  which  is  perpendicular 
to  all  tlie  lines  of  a  plane  passing  through  its  foot  is  said 
to  be  perpendicular  to  the  plane. 

We  have  seen  by  §  107  that  there  may  be  a  straight 
line  perpendicular  to  a  plane,  passing  through  a  point  in 
the  plane.     It  now  remains  to  show  that  there  can  be. 
but  one  such  perpendicular  through  any  point. 

If  PA  be  a  perpendicular,  and  if  another  perpendic- 
ular could  also  be  drawn  through  P 
to  the  plane  MN,  as  PB,  we  should 
have,  by  taking  the  plane  of  the  in- 
tersecting lines,  a  plane  intersecting 
the  given  plane  in  the  line  PC.    PC 

1     •  !•         •        1  ^  -^^--fc-r  •  Fig,  241. 

being  a  line  m  the  plane  MN  is  per- 
pendicular to  both  PA  and  PB.     But  considering  simply 


PERPENDICULARS   TO   PLANES  179 

the  plane  DC,  we  know  from  plane  geometry  that  there 
cannot  be  two  perpendiculars  to  a  line  at  the  same  point. 
Hence  the  supposition  that  a  second  perpendicular  could 
be  drawn  to  the  plane  at  the  point  P  is  not  allowable, 
and  w^e  have  the 

Theorem.  Through  any  point  in  a  j^^cine  one  perpen- 
dicular, and  ONLY  one,  can  be  drawn  to  the  plane. 

Exercises.  —  1.  Prove  that  any  plane  containing  a  perpendic- 
ular to  a  given  plane  will  also  be  perpendicular  to  tlie  given  plane. 

2.  Show  that  if  two  planes  are  perpendicular  to  each  other,  a 
line  perpendicular  to  one  plane  at  a  point  on  the  line  of  intersec- 
tion {trace)  of  the  two  planes  will  lie  in  the  other  plane. 

3.  Show  how  a  carpenter's  square  may  be  used  to  erect  a  per- 
pendicular to  a  plane  at  any  point  in  the  plane. 

4-  Show  that  a  triangle,  the  sides  of  which  are  6,  8,  and  10  (or 
any  multiples  of  0,  8,  and  10)  may  be  used  to  erect  a  perpendicu- 
lar at  a  point  in  a  plane. 

5.  Show  that  through  any  point  not  in  a  plane,  one,  and  only 
one,  perpendicular  can  be  drawn  to  the  plane. 

6.  Establish  the  converse  of  the  theorem  in  this  article. 

7.  Prove  that  the  perpendicular  from  a  point  to  a  plane  is  the 
shortest  distance  from  the  point  to  the  plane. 

8.  Show  that  equal  oblique  lines  will  fall  at 
equal  distances  from  the  foot  of  the  perpendic- 
ular. 

9.  Show  that  the  converse  of  Ex.  8  is  true. 

10.  Show  that  the  opposite  of  Ex.  8  is  true.  Fio.  242. 

11.  Show  how  to  construct  a  perpendicular  to  a  plane  from  a 
point  without  the  plane. 

12.  Show  that  if  two  planes  make  a  right  diedral,  and  a  perpen- 
dicular be  drawn  to  one  plane  from  any  point  in  the  other,  the  foot 
of  the  perpendicular  will  lie  in  the  trace. 


180 


ELEMENTS   OF   GEOMETRY. 


13.  Show  that  through  any  line  in  a  plane,  one  plane,  and  only 
one,  can  be  constructed  perpendicular  to 
the  given  plane. 

14.  Find  the  locus  of  a  point  when 
in  every  position  it  is  equally  distant 
from  three  given  points. 

15.  Find  the  locus  of  a  point  equally 
distant  from  all  points  of  the  circumfer- 
ence of  a  circle.  Fio.  243. 

109.  The  orthographic  projection  of  a  point  on  a  plane 
is  the  foot  of  the  perpendicular  to  the  plane. 

There  may  also  be  oblique  projections.  In  fact,  any 
straight  line  through  the  point,  and  piercing  the  plane, 
will  project  the  point  upon  the  plane  at  the  point  of 
piercing.  But  when  we  use  the  word  projection,  without 
any  modification,  we  mean  the  foot  of  the  perpendicular 
through  the  point  to  the  plane.  This  perpendicular  is 
called  the  projecting  line. 

The  projection  on  a  plane,  of  a  line,  straight  or  curved, 


Fio.  244. 


Fig.  245. 


will  usually  be  a  line,  straight  or  curved.  If  a  moving 
perpendicular  to  the  plane  be  made  to  pass  through 
the  line  that  is  to  be  projected,  the  locus  of  the  foot 
of  the  perpendicidar  will  be  the  projection  of  the  line. 

Theorem.    TJie  projection  of  a  straight  line  on  any  plane 
loill  be  a  straigJU  line. 
Let  JOT"  represent  a  plane,  and  PR  the  straight  line  to 


PROJECTIONS   OF   LINES.  181 

be  projected.  From  any  point,  as  P,  let  fall  the  perpen- 
dicular PF.  The  plane  determined  by  the  two  lines  PR 
and  PF  will  be  the  perpendicular  to  the  plane  MN 
(Ex.  1,  §  108). 

This  plane  (called  the  projecting  plane)  will  contain  all 
points  of  PR,  will  contain  the  projecting  line  in  each 
of  its  positions  (Ex.  2,  §  108),  and  its  trace,  which  is  a 
straight  line,  will  contain  the  feet  of  the  perpendiculars. 

There  will  be  one  projecting  plane,  and  hence  but 
one  projection.  q.  e.  d. 

Corollary.  —  The  converse  is  not 
necessarily  true ;  for  if  the  plane  AB 
make  a  right  diedral  Avith  the  plane 
MN,  any  curve  lying  in  the  plane 
AB  will  be  projected  into  the  trace  j.^^  246. 

AC,  which  is  a  straight  line. 

Only  plane  curves  can  be  projected  in  straight  lines. 

110.  Theorem.  A  straight  line  oblique  to  a  plane  makes 
with  its  projection  both  larger  and  smaller  angles  than  with 
any  other  line  of  the  2)lane  through  the  piercing  point. 

Let  PR  be  any  oblique  line,  RF  its  projection,  and 
RH    any    line    of    the     plane 
through  R. 

Analysis.  —  If  the  Z  PRF  be 
less  than   the  Z  PRH,  and  Ave  • 
should  lay  off  RH  equal  to  RF 
and  draAV  PH,  Ave  would  have 

two  A  PRF  and  PRII  having  tAvo  sides  of  one  equal  to 
two  sides  of  the  other  and  the  included  angles  unequal, 
and  (Ex.  3  at  the  end  of  Chapter  III.)  the  third  side 
PH  Avould  be  greater  than  the  third  side  PF. 


182 


ELEMENTS   OF   GEOMETRY. 


Demonstration.  —  PH  is  greater  than  PF  (being  an 
oblique  line  from  P  to  the  plane  MN).  Hence  the 
Z  PRF  is  less  than  the  Z  PRH. 

But  every  straight  line  that  intersects  another  straight 
line  makes  with  it  adjacent  angles  that  are  supple- 
mentary. 

Hence  Z  PBM>  Z  PBJ.  Q.  e.  d. 

Note. — If  the  analysis  were  omitted,  much  printing  might  be 
saved,  but  the  solution  of  the  problem  would  lose  the  greater  part 
of  its  educational  value. 

111.  Theorem.  If  two  straight  lines  ai'e  perpendicular 
to  the  same  plane,  they  will  be  parallel. 

Let  PR  and  QS  represent  the  perpendiculars. 

The  analysis  suggests  the  follow- 
ing 

Proof.  —  A  plane  through  PR  will 
be  perpendicular  to  MN;  a  plane 
through  QS  will  be  perpendicular  to 
MN.  Each  plane  may  be  so  placed 
that  they  shall  have  the  common  trace  PQ. 
then  coincide  (Ex.  13,  §  108). 

The  lines  PR  and  QS  therefore  lie  in  a  plane ;  and  as 
both  are  perpendicular  to  PQ,  a  line  of  the  plane,  they 
are  parallel.  q.  e.  d. 

Exercises.  —  1.    Establish  the  converse  if  you  can. 

2.  "Will  the  opposite  of  the  theorem  be  true  ? 

3.  Show  that  if  one  of  two  parallels  is  perpendicular  to  a  plane, 
the  other  will  be  also. 

4.  Show  that  if  each  of  two  non-parallel  planes  is  perpendicular 
to  a  third  plane,  their  line  of  intersection  will  be  perpendicular  to 
that  plane. 


Fio.  248. 


They  must 


PERPENDICULAES. 


183, 


112.  Theorem.  If  from  the  foot  of  a  perpendicular  to 
a  plane,  a  line  he  drawn  in  tJie  plane  perpendicular  to  any 
line  of  the  plane,  and  if  its  intersection  with  this  line  be 
joined  with  any  point  in  the  original  perpendicular,  the  last 
line  drawn  will  he  perpendicular  to  the  line  in  the  plane. 


Fig.  2«. 

Let  PR  represent  the  given  perpendicular,  RH  the 
perpendicular  to  AB,  and  PH  the  line  drawn  joining  H 
with  any  point  in  the  perpendicular. 

The  analysis  suggests  the 

Proof.  —  On  AB,  lay  off  equal  distances  from  H  in 
opposite  directions,  to  A  and  B.  Join  A  and  B  with  P 
and  with  R. 


AREA 
A  PR  A 
A  PAH 


ARHB. 
--APRB. 
-APBH. 
=  90°. 


RA  =  RB. 
PA  =  PB. 
Z  PHA  =Z  PHB. 


Q.  E.  D. 


113.  Through  any  point  in  a  plane,  one,  and  only  one, 
parallel  can  be  drawn  to  a  line  of  that  plane.  p.  33. 

If  a  plane  be  rotated  about  any  line  of  the  plane,  and 
the  rotation  be  continued  until  the  plane  returns  to  its 
initial  position,  evei'y  point  in  space  will  have  been 
encountered  once  by  eaxih  part  of  the  rotating  plane. 

It  is  thus  seen  that  one,  and  only  one,  parallel  to  a  Hne 
can  be  drawn  through  any  point  in  space. 


184         ELEMENTS  OF  GEOMETKY. 

Theorem.  If  a  line  not  in  a  given  plane  be  parallel  to 
a  line  of  the  plane,  the  line  not  in  the  plane  will  at  all 
points  he  at  the  same  distance  from  the  plane. 

Let  AB  represent  the  line  D, 

not  in  the  plane,  and  CD  the 
line  in  the  plane  to  which 
AB  is  parallel.  The  analy- 
sis suggests  the  following 

Demonstration.— Dtqw  AO  Fig.  250. 

and  BD,  perpendiculars  to  the  two  parallels.     In  the 
plane  MN  draw  CE  and  DFl.  to  CD  at  C  and  D. 

The  plane  determined  by  AC  and  CE  will  be  perpen- 
dicular to  CD  and  AB.  So  with  the  plane  determined , 
by  BD  and  DF. 

From  A  and  B,  perpendiculars  to  the  plane,  MN  will 
fall  in  CE  and  DF. 

A  ACE  =  A  BDF, 

since  AC  =  BD,  ZACE  =  ZBDF,  and  each  are  right 

triangles. 

.-.  AE  =  BF.  Q.E.D. 

Note.  — The  line  AB  is  said  to  be  parallel  to  the  plane. 

Ezercises.  —  1.  Show  that  if  a  line  EF  be  dravra,  it  will  be 
parallel  to  CD. 

2.  Show  that  any  plane  through  AB  will  intersect  the  plane 
MN  (if  it  intersect  it  at  all)  in  a  line  parallel  to  AB. 

8.  Show  that  if  two  lines  are  parallel  to  a  plane  and  at  the  same 
distance  from  it,  they  will  lie  in  the  same  plane,  which  will  have 
all  its  points  equally  distant  from  the  given  plane. 

Note.  — The  two  planes  hi  Ex.  3  are  called  parallel.  They  are 
said  to  never  intersect.  Also  they  are  sometimes  said  to  intersect 
in  a  line  at  infinity.     Both  statements  mean  the  same  thing. 


PROBLEMS. 


185 


114.    Problems.  —  1.    Show  that  if   two  parallel  planes  are 
intersected  by  a  third  plane,  the  lines  of 
intersection  will  be  parallel. 

2.  Show  that  if  one  of  any  number  of 
parallel  planes  be  cut  by  a  plane,  all  will 
be,  and  the  lines  of  intersection  will  be 
parallel. 

3.  Show  that  two  or  more  lines 
which  pierce  three  parallel  planes  will 
be  divided  proportionally. 


Fig.  251. 


Aiib' 


L 


^^ 


/ 


7 


i~\ — \ 


7 


Fig.  252. 


4.  Show  that  the  projecting  line  of  the  middle  point  of  a 
segment  that  is  projected  on  a  plane  will  be  the  half-sum  of  the 
projecting  lines  of  the  segment  ex- 
tremities. 

5.  If  from  any  point  within  a  die- 
dral,  perpendiculars  be  let  fall  on  the 
planes  forming  the  diedral,  the  angle 
formed  by  the  perpendiculars  will  be 
the  supplement  of  the  diedral.*  V/    fig.  253. 

6.  Show  that  a  plane  may  be  passed  through  a  line  parallel  to 
a  given  line ;  and  that  a  plane  may  be  passed  through  a  point 
parallel  to  two  given  lines. 


*  "When  two  planes  intersect,  they  make  angles  that  are  supple- 
mentary to  each  other,  just  as  two  intersecting  lines  do. 


186  ELEMENTS    OF   GE0:METRY. 

7.  Two  lines  in  space  which  are  not  parallel  may  be  joined  by 
one,  and  only  one,  mutual  perpendicular ;  and  the  perpendicular 
will  be  shorter  than  any  other  distance  between  points  on  the 
two  lines. 

Let  AB  and  CD  represent  the  two  lines. 

(a)  Analysis.  —  If    the    lines    have    a 
mutual  perpendicular,  and  FB  represent 
it,  it  would  be  perpendicular  to  AB  and 
to  BK—a.  line  through  B\\to  CD  — and  Fig.  2&i. 
therefore  perpendicular  to  the  plane  of  AB  and  BK.     (Let  MN 
represent  this  plane,  and  BK  the  projection  of  CD  on  the  plane.) 

Demonstration.  — Through  one  of  the  lines  pass  a  plane  parallel 
to  the  second  line.  Project  the  second  line  upon  this  plane,  and 
at  its  point  of  intersection  with  AB  erect  a  perpendicular  to  the 
plane.  Being  perpendicular  to  the  plane,  it  will  be  perpendicular 
to  AB  and  BK.  And  being  perpendicular  to  BK,  it  will  be  per- 
pendicular to  CD.  Hence  a  perpendicular  can  be  drawn  joining 
two  lines  in  space. 

(b)  Analysis. — If  a  second  mutual  perpendicular  could  be 
drawn,  and  FE  represent  it,  a  line  FT,  through  F II  to  CD,  would 
also  be  a  perpendicular  to  FEy  and  the  plane  of  the  lines  AB  and 
FT  will  be  perpendicular  to  FE.  But  the  plane  of  the  lines  AB 
and  FT  would  be  the  same  as  the  plane  of  the  lines  AB  and  BK. 
Hence  FE  would  be  perpendicular  to  MX,  and  J^  would  be  the 
projection  of  the  point  E  upon  the  plane  MX. 

Demonstration.  —  Thi'ough  one  of  the  lines  pass  a  plane  parallel 
to  the  second  line.     Project  the  second  line  upon  the  plane. 

Our  analysis  shows  that  if  FE  be  a  mutual  perpendicular  to 
AB  and  CD,  the  line  FT  must  be  the  projection  of  CD.  But  by 
§  109,  a  straight  line  can  have  but  one  projection  on  a  given  plane. 
Therefore  FE  cannot  be  a  mutual  perpendicular,  and  there  can 
be  but  one  mutual  perpendicular  to  two  lines  in  space. 

Discuss  the  problem  —  including  the  cases  when  the  two  lines 
are  parallel  and  when  they  intersect. 

8.  Show  that  if  a  line  be  perpendicular  to  one  of  two  inter- 
secting planes,  its  projection  upon  the  other  plane  will  be  per- 
pendicular to  the  line  of  intersection  of  the  two  planes. 


PROBLEMS. 


187 


9.  Show  that  if  two  angles  in  space  have  their  sides  parallel 
and  extending  in  the  same  direction, 
the  angles  will  be  equal  and  the  planes 
of  the  angles  will  be  parallel. 

10.  Show  that  the  locus  of  a  point, 
the  ratio  of  the  distances  of  which  from 
two  given  planes,  is  ±1,  will  be  the  Fig.  255. 

plane  bisectors  of  the  diedral  angles,  and  will  themselves  form 
a  right  diedral. 


CHAPTER  X. 


115,  Definitions.  1.  The  figure  formed  by  a  surface,  all 
points  of  which  are  equally  distant  from  a  fixed  point, 
is  called  a  sphere. 

2.  The  fixed  point  is  called  the  centre. 

3.  The  surface  completely  encloses  a  portion  of  space. 

4.  The  surface  is  the  locus  of  a  point  at  a  fixed  dis- 
tance from  a  given  point. 

Theorem.     Every  plane  section  of  a  sphere  is  a  circle. 
Let  C  represent  the  centre  of  the  sphere,  and  P,  Q, 
and  B  points  in  the  section. 

Analysis.  —  If  the  plane 
section  be  a  circle,  it  Avill 
have  a  centre,  and  if  at  that 
centre  a  perpendicular  to  the 
plane  be  erected,  it  will  pass 
through  the  centre  of  the 
sphere  (see  Ex.  15,  §  108). 

Demonstrat  ion.  —  From  C 
let  fall  a  perpendicular  CF  upon  the  plane.  Join  P  with 
F  and  C.  Kotate  the  right  A  CPF  about  Ci^as  an  axis. 
The  point  P  as  it  moves  will  be  at  a  fixed  distance  from 
C,  and  so  must  be  on  the  surface  of  the  sphere ;  it  will 
also  be  in  the  secant  plane  because  FP  will  be.  The 
point  P  will  therefore  move  over  the  line  of  intersection 
of  the  two  surfaces. 

188 


Fig.  256. 


SPHERICAL   ARCS.  189 

But  the  locus  of  a  point  in  a  plane  at  a  given  distance 
from  a  fixed  point  is  the  circumference  of  a  circle,   q.  e.  d. 

Definitions.  1.  The  points  iV  and  S,  where  the  line  CF 
pierces  the  surface  of  the  sphere,  are  called  poles  of  the 
circle  PQR,  and  the  line  NS  is  called  its  axis. 

2.  If  the  secant  plane  pass  through  the  centre  of  the 
sphere,  the  circle  of  intersection  is  called  a  great  circle — 
otherwise,  a  small  circle. 

Exercises.  —  1.  Show  that  all  great  circles  of  a  sphere  are 
equal,  and  that  the  intersection  of  any  two  will  be  a  diameter  of 
the  sphere. 

2.  Show  that  a  plane  section  not  through  the  centre  of  a  sphere 
will  determine  a  circle  of  smaller  radius  than  a  great  circle. 

3.  Show  that  the  greater  the  distance  from  the  centre  that  the 
secant  plane  is  passed,  the  smaller  will  be  the  circle  of  inter- 
section. 

4.  Show  that  if  a  circle  be  revolved  about  a  diameter  as  an  axis, 
it  will  generate  a  sphere. 

5.  Show  that  the  pole  N  or  S  is  equally  distant  from  all  points 
of  the  circumference  of  the  circle  of  intersection. 

6.  Show  that  in  general,  one,  and  only  one,  great  arc  may  be 
passed  through  two  points  on  the  surface  of  a  sphere. 

7.  Show  that,  in  general,  an  infinite  number  of  arcs  of  small 
circles  may  be  passed  through  two  points  on  the  surface  of  a 
sphere.  ^^ 

8.  Making  use  of  the  facts  determined  in  /  i 
Ex.  5,  §  10-3,  show  that  the  arc  of  a  great  /  j 
circle  joining  two  points  on  the  surface  of  a  [  i 
sphere  will  be  shorter  than  the  arc  of  a  \  ', 
small  circle  joining  them.                                     \         \ 

JRemark.— Thin  fact  plays  an  important  \^^ 
part  in  ocean  navigation.  ^ig  25" 

9.  An  arc  of  a  great  circle  which  subtends  an  angle  of  90° 


190         ELEMENTS  OP  GEOMETRY. 

is  called  a  quadrant.    Show  that  all  points  of  a  great  circle  are  at 
a  quadrant's  distance  from  its  poles. 

10.  A  plane  through  its  centre  bisects  the  surface  of  a  sphere 
and  also  its  volume. 

116.  Definition.  The  angle  between  any  curves  which 
intersect  is  the  angle  made  by  the  tangents  to  the  curves 
at  the  point  of  intersection. 

Let  two  great  circles  intersect  on  the  diameter  AX. 
Through  the  centre  pass  a  plane 
perpendicular  to  AX-,   and  at  A 
draw  tangents  to  the  great  circles. 
Z  HCK=  Z  H-AX-K, 
Z  NAM=  Z  KCH=  KH. 

The  angle  between  the  arcs  HA 
and  KA  is  said  to  be  measured  by 
the  arc  KH,  of  a  great  circle 
included  between  its  sides  and 
90°  from  the  vertex.  f,q  258. 

Ezercises.  —  1.  Show  that  the  poles  of  all  great  circles  which 
pass  through  AX  will  be  found  in  the  circumference  KHQ,  the 
plane  of  which  is  perpendicular  to  ^Xat  its  middle  point. 

2.  Show  how,  at  a  given  point  of  an  arc,  to  erect  an  arc  perpen- 
dicular to  the  given  arc. 

3.  Through  any  two  points  show  how  to  pass  the  arc  of  a  great 
circle. 

4.  Show  how,  through  a  point  on  a  sphere,  to  let  fall  a  perpen- 
dicular to  a  given  great  circle. 

Definitions.  —  1.  Circles  having  the  same  poles  are  said  to  be 
parallel. 

2.  The  portion  of  the  surface  of  a  sphere  included  between 
parallel  planes  is  called  a  zone. 

3.  The  volume  corresponding  to  a  zone  is  called  a  spherical  seg- 
ment. 


SPHERICAL   ARCS. 


191 


Fig.  259. 


Note.  —  A  spherical  blackboard  is  a  most  useful  adjunct  to  the 
study  of  spherical  geometry.  A 
diameter  of  20  inches  is  a  con- 
venient size.  A  quadrant  made 
of  wood  is  also  a  great  con- 
venience. If  a  regular  spherical 
blackboard  of  the  size  suggested 
cannot  conveniently  be  had,  a 
papier-machfi  globe  can  be  slated 
over  and  made  to  serve  the  purpose  fairly  well. 

5.  Show  that  if  a  great  circle  and  a  small  circle  have  the  same 
poles,  they  will  intercept  equal  arcs  of  the 
great  circles  which  pass  through  the  poles. 

6.  Show  that  two  small  circles  having 
the  same  poles  will  intercept  equal  arcs 
of  the  great  circles  which  pass  through 
the  poles. 

7.  Show  that  if  two  great  circles  inter- 
sect as  in  AX,  and  from  points  of  one  of 
them  perpendiculars  be  let  fall  to  the 
other,  they  will  vary  in  length  from  0  to  ^  ***'  '^'^^' 

the  measure  of  the  angle,  as  the  point  from  which  the  perpendic- 
ulars are  let  fall  passes  from  AtoH. 

8.  If  AB  is  tangent  to  a  given  great  circle  ETD  at  T,  and  if 
through  AB  any  plane  be  passed,  it  will,  in  general,  intersect  the 


Pig.  261.  Fia.  262. 

sphere  in  a  small  circle,  which  will  be  tangent  to  the  great  circle  at 


192 


ELEIkfENTS   OF   GEOMETRY. 


Fia.  263. 


r,  will  lie  entirely  in  one  of  the  hemispheres  determined  by  the 
great  circle,  and  will  not  touch  the  circumference  of  the  great 
circle  in  any  other  point. 

Demonstration. — The  only  portion  of  the  planes  of  the  two 
circles  which  are  in  common,  is  the  line  AB,  of  which  the  point  T 
is  the  only  point  on  the  surface  of  the  sphere.  AB  being  in  the 
plane  of  each  circle,  and  touching  each  at  but  one  point,  is  tangent 
to  each  at  that  point.  Hence  the  circles  are  tangent  to  each  other 
at  the  same  point  and  it  is  the  only  point  common  to  both. 

117.   Definitions.     The  two  equal  parts  (Ex.  10,  §  115) 
into  which  the  circumference  of  a  great  circle 
separates  the  surface  of  a  sphere,  are  called 
hemispheres. 

The   circumferences   of    two   great   circles 
separate  the  surface  of  a  sphere  into  four 
parts,  each  of  which  is  called  a  lune.     The  surface  in- 
cluded by  the  semicircles  AED  and  ABD  is  a  lime. 

The  circumferences  of  three  great  cir- 
cles, when  they  do  not  have  a  common 
diameter,  separate  the  surface  into  eight 
parts,  each  of  which  is  called  a  spherical 
triangle  (sph.  A). 

If  through  the  vertices  (A,  B,  and  C) 
of  a  spherical  triangle,  diameters  be 
drawn,  their  other  extremities  (P,  Q, 
and  R)  determine  a  spherical  triangle 
which  is  called  the  symmetrical  of  the 
first  spherical  triangle. 

Theorem.  —  T7ie  angles  and  skies  of 
a  spherical  triangle  and  its  symmetrical 
are  mutually  equal. 

Let  ABD  represent  a  spherical  triangle,  and  E,  F,  G, 


Fir,.  264. 


Fig.  265. 


SPHERICAL  TRIANGLES. 


193 


the  extremities  of  the  diameters  of  the  sphere  through 
A,  B,  and  D.     The  sph.  AEFG  is 
the  symmetrical  of  the  sph.  A  ABD. 

/.  BAD  =  Z.  GAF 

(being  vertical) ; 

ZGAF=ZGEF 

(being  the  angle  between  the  planes 

of  AFE  and  AGE). 

.-.  ZBAD  =  ZGEF. 


FiQ.  266. 


In  the  same  way  we  may  show  that 

Z  ADB  =  Z  FGE 

and  Z  ABD  =  Z  EFG. 

AB  =  EF  (each  being  the  supplement  of  AF), 

AD  =  EG  (each  being  the  supplement  of  DE), 

and   BD  =  FG  (each  being  the  supplement  of  DF). 

Q.  E.  D. 

Exercise.  —  1.  Show  that  if  a  spherical  triangle  is  isosceles,  it 
may  be  brought  to  coincide  with  its  symmetrical. 

Note.  —  In  general  the  spherical  triangles  that  are  symmetrical 
cannot  be  superimposed,  because  the  equal  parts  are  not  arranged 
in  the  same  order,  and  if  one  figure  be  reversed,  the  curvature  will 
not  permit  of  coincidence  of  surface. 

The  plane  triangle  that  has  its  parts  equal,  but  differently 
arranged,  from  another  triangle,  may  be  reversed  and  the  two 
brought  to  coincide. 

The  fact  that  symmetrical  spherical  triangles  are  equivalent  in 
area  will  be  established  later.  At  present  we  are  only  concerned 
with  the  facts  presented  in  the  theorem. 


194 


ELEMENTS   OF   GEOMETRY. 


118.  Theorem.  —  If  at  the  middle  point  of  a  given  seg- 
ment of  a  great  circumference,  a  perpendicular  great  arc  be 
erected,  all  points  in  the  perpendic- 
ular arc  vjill  he  equally  distant  from 
the  extremities  of  the  segment. 

Let  AB  represent  the  arc  seg- 
ment, of  which  M  is  the  middle 
point.  Let  MQP  represent  the 
perpendicidar  arc,  and  Q  any  point 

of  i*-  Fig.  267. 

Analysis.  —  If  QB=  QA,  the  sph.  A  QMB  and  QMA 
would  have  the  three  sides  of  one  equal  to  the  three  sides 
of  the  other,  and  the  angles  at  M  in  each  equal,  but  we  are 
unable  to  substitute  the  one  for  the  other  and  show  that 
they  are  identical.  This  suggests  the  use  of  the  sym- 
metrical spherical  triangle.     Hence  the 

Demonstration.  — The.  diameters  through  Q,  M,  and  B 
determine  the  symmetrical  sph.  A  Q'M^B'.  Then  if  the 
symmetrical  sph.  A  Q'WB'  be  caused  to  slide  over 
the  surface  of  the  sphere,  rotating  180°  about  the  axis  of 
the  Q  MQP,       ^,  ^.jj  coincide  with  MQ, 


the  rt.  Z  Q'WB'     " 

JifB'      " 


«  Z  QMA, 

"   m1. 


and  Q'B'      ".         "  «    QA 

(because  between  two  points  on  the  surface  of  a  sphere 
one,  and  only  one,  great  arc  can  be  drawn). 

But  Q^'  =  QB. 

.-.  (^=qB.  9.E.D, 


SPHERICAL   ARCS. 


195 


119.  Theorem. — From  any  point  on  the  surface  of  a 
sphere,  the  perpendicular  arc  (less  than  90°)  to  any  great 
circle  will  be  less  than  any  oblique  arc. 

The  analysis  suggests  that  if  with  Q  as  a  pole  and  QM 
as  an  arc  radius,  a  small  circle  be 
constructed  on  the  surface  of  the 
sphere,  it  will  be  tangent  to  the 
great  circle  at  M  and  will  have 
no  other  point  in  common  with  it 
(Ex.  8,  §  116);*  and  as  the  arc- 
distance  from  Q  of  every  point 
in  the  circumference  of  the  small 
circle  is  the  same  as  QM, 


Fig.  268. 


QM=  (ff^<  QB. 


Q.  E.  D. 


Exercises.  —  1.  Show  that  QPS  will  be  greater  than  any 
oblique  arc  from  Q  to  the  great  circle  ABH. 

2.  Show  that  of  two  oblique  arcs  from  Q  to  the  great  circle 
ABH,  the  one  that  falls  at  the  greater  distance  from  M  will  be 
the  greater. 

3.  Show  that  as  the  oblique  arcs  increase,  the  angle  made 
by  them  with  the  base  of  the  hemisphere  will  decrease  from  90° 
until  the  oblique  arc  shall  have  reached  90°  (when  the  angle  will 
be  measured  by  the  arc  MQ) ,  and  that  thereafter  the  angle  will 
increase  until  the  perpendicular  arc,  QPH,  is  reached. 


*NoTE.  —  Any  chord  of  a  sphere,  except  a  diameter,  may  be  the 
common  chord  of  a  great  circle  and  any  number  of  small  circles  or 
of  two  or  more  small  circles.  A  small  circle  may  be  tangent  to 
another  small  circle  or  to  a  great  circle.  But  a  great  circle  cannot 
be  tangent  to  another  great  circle. 

The  student  may  be  aided  in  his  conception  of  these  relations, 
by  passing  a  plane  through  the  diameter  or  chord  under  con- 
sideration and  rotating  it,  observing  the  spherical  sections. 


196 


ELEMENTS   OF   GEOMETRY. 


Fig.  269. 


120.  Theorem.  —  The  sum  of  two  sides  of  a  spherical 
triangle  is  greater  than  the  third  side. 

Let  ABC  represent  tlie  spherical  triangle,  AC  being 
the  largest  side. 

Analysis.  —  If  A  Che  less  than 

AB  4-  BO,  it  may  be  separated 
into  two  parts,  one  of  which  shall 

be  less  than  AB  and  the  other 

less  than  CB.     Can  this  be  done  ? 

The  student  will  furnish  the  demonstration. 

Exercises.  —  1.  Show  that  the  difference  between  two  sides  of 
a  spherical  triangle  will  be  less  than  the  third  side. 

2.  It  has  been  shown  in  §  118  that  all  points  in  the  perpendic- 
ular arc  bisecting  a  given  segment  will  be  equally  distant  from  the 
extremities.  Let  the  student  now  show  that  any  point  not  on 
such  perpendicular  will  not  be  equally  distant  from  the  segment 
extremities,  but  will  be  nearer  the  extremity  on  its  own  side  of 
the  perpendicular. 

Note.  —  The  student  will  observe  the  similarity  of  method  and 
relation  in  Plane  and  Spherical  Geometry. 

121.  Problems.  —  1.  To  construct  a  small  circle  the  circumfer- 
ence of  which  shall  pass  through  the  vertices  of  a  spherical  triangle. 

2.  Making  use  of  Ex.  1,  §  117  and  Prob. 
1  above,  show  that  two  symmetrical  spher- 
ical triangles  are  equivalent  in  area. 

3.  Show  that  if  the  three  sides  of  a  spher- 
ical triangle  be  given,  the  spherical  triangle 
may  be  constructed. 

It  follows  that  if  two  spherical  triangles 
have  the  three  sides  of  one  equal  to  the  three 
sides  of  the  other,  the  spherical  triangles  will 
be  superimposable  or  symmetrical,  and  so  will  have  the  angles  as 
well  as  sides  equal,  and  will  be  equivalent  in  area. 


Fig.  270. 


SPHERICAL   TRIANGLES. 


197 


Fig.  271. 


4.  To  construct  on  the  surface  of  a  sphere  an  angle  equal  to 
a  given  spherical  angle. 

Let  Z  ABD  represent  the  given  spherical 
angle,  and  P  the  point  on  a  given  arc  QP, 
where  the  vertex  of  the  required  angle  is  -^f 
to  be  drawn. 

The  analysis  suggests  that  if  any  arc 
AD  be  drawn,  a  distance  PQ,  equal  to  JBA, 
be  laid  off  from  P,  either  with  dividers  or 
tape,  and  on  that  as  a  base,  a  spherical 
triangle  with  sides  equal  to  those  of  sph.  AADB  be  constructed 
(after  the  manner  indicated  in  Prob.  3),  the  Z  QPK  will  equal 
the  Z  ABD. 

Bemark.  —  It  is  customary  —  and  convenient  —  to  lay  off  BA 
and  BD  equal ;  but  the  principle  is  of  course  the  same. 

5.  Show  that  a  spherical  triangle  will  be  less  than  a  hemisphere 
in  area. 

Note. — No  side  or  angle  of  a  spherical 
triangle  may  be  greater  than  180°.  If  a 
three-sided  figure  should  be  constructed  on  a 
sphere,  one  side  of  which  is  greater  than 
180°,  the  remainder  of  the  hemisphere  is 
regarded  as  the  spherical  triangle. 

If  a  three-sided  figure,  one  angle  of  which 
is  greater  than  180°,  be  constructed  on  a 
sphere,  the  arcs  will  all  lie  on  a  hemisphere  and  the  excluded  part 
of  the  hemisphere  is  regarded  as  the  spher- 
ical triangle,  no  angle  of  which  will  be  180°. 

6.  Show  how  to  construct  a  spherical 
triangle,  having  given  two  sides  and  the 
included  angle. 

Note.  —  It  follows  that  if  two  spherical 
triangles  have  two  sides  and  the  included 
angle  of  one  equal  to  two  sides  and  the 
included  angle  of  the  other,  they  will  be 
superimposable,  or  symmetrical,  and  hence  have  the  other  side  and 
remaining  angles  equal,  and  areas  equivalent. 


Fig.  2T2. 


Fig.  273. 


198  ELEMENTS   OF   GEOMETRY. 

7.  Show  how  to  construct  a  spherical  triangle,  having  given  two 
angles  and  the  included  side. 

Note.  —  It  follows  that  two  spherical  triangles  having  two  angles 
and  included  side  of  one  equal  to  two  angles  and  included  side  of 
the  other,  will  have  their  remaining  parts  equal  and  will  be  equiv- 
alent in  area. 

8.  To  construct  a  spherical  triangle, 
having  given  two  sides  and  an  angle 
opposite  one  of  them. 

At  any  point  (A)  of  a  great  arc,  con- 
struct the  given  angle.     On  either  side  of 

the  angle,  lay  off  the  given  adjacent  side 

/ — \ 
(AB).     With  an  arc-radius  equal  to  the 

Vie   274 

side  that  is  to  be  opposite  the  given  angle, 

and  with  J5  as  a  pole,  construct  a  small  circle  intersecting  AG  a,t 
D  and  D'.  Joining  them  with  the  point  J5,  we  will  have  the  two 
sph.  A  ADB  and  AD'B,  that  contain  the  given  parts. 

Discussion.  — If  from  ^  a  ±  BF  be  let  fall  on  the  AG  (Ex.  4, 
§  116),  it  will  be  the  shortest  distance  from  B  to  AG  (§  119). 

(a,  1.)  If  the  side  opposite  the  given  angle  be  less  than  the 
perpendicular,  there  will  be  no  construction. 

(a,  2.)  If  equal  to  the  perpendicular,  there  will  be  one  con- 
struction —  the  spherical  triangle  being  right-angled. 

(a,  3. )  If  intermediate  in  value  between  the  perpendicular  and 
BA,  and  also  intermediate  in  value  between  the  perpendicular  and 
BA'  (the  supplement  of  BA),  there  will  be  tvoo  constructions. 

(a,  4.)  If  intermediate  in  value  between  either  of  the  arcs, 
AB  or  A'B,  and  the  perpendicular,  but  not  between  the  other  one 
and  the  perpendicular,  there  will  be  but  one  construction. 

(a,  6.)  If  not  intermediate  in  value  between  the  perpendicular 
and  either  AB  or  A'B,  there  will  be  no  construction. 

What  we  have  seen  in  the  discussion  so  far  has  been  developed 
under  the  supposition  that  the  ZBAG  was  an  acute  angle.  Let 
as  consider  the  case  where  it  is  obtuse. 


SPHERICAL   TRIANGLES. 


199 


Fig.  275. 


The  perpendicular  from  J5  will  pass 
through.  P  (the  pole  of  AG),  will  be 
greater  than  90°,  and  will  be  greater 
than  any  other  arc  drawn  from  B  to  any 
point  in  AG. 

(b,  1.)  If  the  side  opposite  the  given 
angle  be  greater  than  this  perpendicular, 
there  will  be  no  construction. 

(&,  2.)  If  the  side  be  equal  to  the  perpendicular,  there  will  be 
one  consti'uction,  and  the  spherical  triangle  will  be  right-angled. 

(6,  3.)  If  the  side  be  intermediate  in  value  between  the  perpen- 
dicular and  BA,  and  also  intermediate  between  the  perpendicular 
and  BA',  there  will  be  two  constructions. 

(6,  4.)  If  the  side  be  intermediate  in  value  between  the  perpen- 
dicular and  but  one  of  the  arcs  AB  or  A'B,  there  will  be  one  con- 
struction only. 

(6,  5.)  If  not  intermediate  in  value  between  the  perpendicular 
and  either  the  other  given  side  or  its  supplement,  there  will  be  no 
solution. 

Note.  —  Case  (a,  1)  might  be  included  with  (a,  5),  and  (6,  1) 
with  (6,  5)  ;  but  for  the  sake  of  gradual  approach  it  is  deemed 
best  to  present  them  in  this  order. 

122.  Definitions.  If  with  the  vertices  of  a  spherical  tri- 
angle as  poles  and  quadrants  as  arc-radii,  a  spherical 
triangle  be  constructed,  it  is  called  the 
polar  of  the  given  spherical  triangle. 

Let  ABC  represent  the  given  spher- 
ical triangle,  and  DK,  KG,  and  GD, 
the  arcs  constructed  with  B,  A,  and  C 
as  poles. 

Since  all  points  of  DK  are  at  a 
quadrant's  distance  from  B,  and  all 

points  of  GD  are  at  the  same  distance  from  C,  D  will  be 
the  pole  of  BG. 


200         ELEMENTS  OF  GEOMETRY. 

For  the  same  reasons,  K  will  be  the  pole  of  AB,  and 
G  will  be  the  pole  of  AC. 

Thus  it  is  seen  that  the  relation  is  a  mutual  one,  and 
if  sph.  A  DKO  is  the  polar  of  ABC,  the  latter  is  also  the 
polar  of  the  former. 

I)M=  90°, 

.'.  DM-\-NI<:=im°, 

or  DN+2  NM  +  3/7r  =  180°, 

or  I>^+^^=180°. 

But  NM  =  ZB.  (§  116) 

.'.  DK -{- Z  B  =  1S0°. 

In  the  same  way  it  may  be  shown  that : 

I^  +  ZA  =  180°, 

and  Gb  +  ZC=  180°. 


Again,  EC  =  90°, 

^=90°. 
.-.  EC  +  AJ  =  180°  =  EA. -\- 2  AC -\- C J, 
or  ^+^'0=180°. 

But  EJ=ZG. 

.-.  ZG  + AC  =180". 
In  the  same  way  it  may  be  shown  that  BC  +  ZD  = 
180°,  and  52  +  Z  A"  =  180°.     Hence  the 


SPHERICAL   TRIANGLES.  201 

Theorem.  If  tivo  spherical  triangles  are  polar  to  each 
other,  any  side  of  either  will  he  the  supplement  of  the  angle 
in  the  other  at  the  pole  of  that  side. 


Fig.  277. 

Note.  —  If  any  of  the  sides  of  a  spherical  triangle,  or  its  polar, 
are  greater  than  90*^,  some  of  the  sides  of  one  will  cross  sides  of 
the  other. 

123.  Problems.  —  1.  To  construct  a  spherical  triangle,  hav- 
ing given  the  three  angles. 

Analysis.  —  If  we  had  the  required  spherical  triangle  and  should 
construct  its  polar,  the  sides  of  the  latter  would  be  the  supple- 
ments of  the  three  given  angles. 

Construction.  —  With  the  supplement*  of  each  of  the  given 
angles  as  sides,  construct  a  spherical  triangle  (Ex.  3,  §  121).  This 
will  be  the  polar  of  the  required  spherical  triangle.  Construct  its 
polar ;  it  will  be  the  required  spherical  triangle. 

2.  To  construct  a  spherical  triangle,  having  given  two  angles  and 
the  side  opposite  one  of  them,  making  use  of  the  polar  spherical 
triangle. 

3.  To  construct  a  spherical  triangle,  having  given  two  angles  and 
the  side  opposite  one  of  them,  without  using  the  polar.  Make  a 
careful  discussion. 

4.  Show  that  0°  and  360°  are  the  limits  of  the  sum  of  the  three 
sides  of  a  spherical  triangle. 

*  To  get  an  arc  the  supplement  of  an  angle,  take  that  part  of  a 
semicircumference  having  the  vertex  of  the  angle  as  its  pole,  not 
included  by  the  arcs  forming  the  angle. 


202         ELEMENTS  OF  GEOMETRY. 

5.  Show  that  180°  and  540°  are  the  limits  of  the  sum  of  the 
three  angles  of  a  triangle. 

6.  The  student  will  observe  that  when  the  three  angles  are 
given  of  a  plane  triangle,  the  triangle  is  midetermined  (Ex.  6,  §  28), 
while  the  three  angles  of  a  spherical  triangle  completely  determine 
it.  Show  from  this  that  if  two  spherical  triangles  have  the  three 
angles  of  each  equal,  each  to  each,  the  spherical  triangles  are 
equal  or  symmetrical. 

124.  Theorem  1.  If  two  sides  of  a  spherical  triangle 
are  equal,  the  angles  opposite  them  will  he  equal. 

Let  BA  and  BC  represent  the  equal 
sides. 

From  the  middle  point  M  of  the  arc 
AC  draw  the  arc  MB.  A- 

AAMB  =  A  CMB  (3  sides  equal). 

.*.   /.A=^ZC,  Q.  E.  D. 

Z  AMB  =  Z  CMB  =  90% 
Z  ABM=  Z  CBM. 

Theorem  2.  Conversely,  if  two  angles  of  a  spherical 
triangle  are  equal,  the  sides  opposite  them  will  be  equal. 

If  the  sides  are  not  equal,  and  AB  >  BC,  a  perpendic- 
idar  erected  at  the  middle  point  of 
the  side  included  by  the  equal 
angles  will  not  pass  through  the 
vertex.  (The  perpendicular  is  the 
locus  of  all  points  equally  distant 
from  A  and  C.)  It  will  therefore 
intersect  the  sides  in  succession  as 
at  H  and  K.  If  the  auxiliary  HC  be  drawn,  it  will 
make  with  MC  an  angle  equal  to  the  angle  at  A,  but 


SPHERICAL  TRIANGLES.  203 

smaller  than  the  Z  BCA.     But  that  is  contrary  to  the 
hypothesis. 

Hence  the  two  sides  cannot  be  unequal;  or  in  other 
words,  they  are  equal.  q.  e.  d. 

Exercises. — 1.  Construct  a  right  spherical  triangle,  having 
given  the  sides  adjacent  to  the  right  angle. 

2.   The  same,  having  given  the  hypotlienuse  and  one  side. 

8.  The  same,  having  given  the  hypothenuse  and  an  oblique 
angle. 

4.  The  same,  having  given  an  oblique  angle  and  the  adjacent 
base. 

6.  The  same,  having  given  an  oblique  angle  and  the  side 
opposite.    Discuss  thoroughly. 

6.  Show  that  two  of  the  angles  of  a  spherical  triangle  may 
each  be  90^  ;  and  that  when  they  are  such,  the  sides  opposite  them 
will  be  90°. 

7.  Show  that  the  three  angles  of  a  spherical  triangle  may  each 
be  90°  ;  that  when  they  are  such,  each  side  of  the  spherical  triangle 
will  be  90° ;  and  that  the  area  of  the  spherical  triangle  will  be 
one-eighth  of  the  surface  of  the  sphere. 

8.  Show  that  if  two  spherical  triangles  can  be  brought  to  have 
a  common  side,   and  two   other   sides  cross            „         p 
each  other,  the  sum  of  the  sides  that  cross                     "' 
is  greater  than  the  sum  of  the  sides  that  do 
not  cross  ;  i.e.  -^£ 3k.B 

AC+BE>AB+  CE.  ^'•^-  ^^■ 

9.  Show  that  if  two  spherical  triangles  have  a  common  side  AC, 
and  the  vertex  Z>,  of  one,  lies  within  the  other,  ^ 

AB  +  BC>AI>  +  DC.  X^^ 

10.  Show  that  if  two  spherical  triangles 
have  two  sides  in  one  equal  to  two  sides  in  -^ 
the  other,  and  the  included  angles  are  un- 
equal, the  third  sides  will  be  unequal,  and  the  greater  side  will  be 
opposite  the  greater  angle. 


Fig.  281. 


204 


EliEMENTS   OF   GEOMETRY. 


Fig.  2S8. 


11.    Show  that  the  sum  of  the  sides  of  a  convex*  spherical 
polygon  of  any  number  of  sides  is  less  than  a  great  circle. 

Let  ABODE  represent  any  spher- 
ical polygon.  Continue  any  side,  as 
AB,  so  as  to  form  a  great  circle ;  it 
separates  the  sphere  into  hemispheres. 
The  angles  at  A  and  B  are  less  than 
180°  each.  If  any  of  the  vertices  of 
the  polygon  should  fall  outside  of  the 
hemisphere  in  which  AE  and  JBClie, 
we  should  have  some  of  the  angles 
greater  than  180°.  And  in  case  the 
polygon  were  separated  into  spherical  triangles  by  means  of  diag- 
onals, some  of  the  spherical  triangles  would  be  exterior  to  the 
polygon,  and  the  polygon  would  not  be  convex.  Therefore  the 
spherical  polygon  lies  entirely  within  the  hemisphere. 

Produce  AE  to  A',  ED  to  H,  and  DC  to  K. 
aFa'  =  AEA'  =  AE+  EA', 
EA'  +  jVH>ED  +  DH, 

Dn+Hk>DC+  ck, 
ck-\-  kb>6b, 

AB=BA, 
AFA>  +  EA<  +  A^H -\- DH  +  nk  +  CK  +  KB  + AB>AE+ EA' 

-\-  ED  +DH+DC+  Ck+  CB  +  BA; 
AFA'  +  A^H  +  HE  +  KB  +  AB  >  AE -^  Fd  +  DC  +  CB  +BA; 
Gre&t  circle  >  AE  +  ED  +  DC  +  CB+BA. 

Q.E.D. 

•Note.  — A  convex  spherical  polygon  is  one  in  which  none  of 
the  angles  are  greater  than  180°. 


4 


CHAPTER  XI. 


125.  Definitions.  We  have  seen  that  a  single  plane  sep- 
arates space  into  two  parts;  two  planes  separate  it  into 
four  parts,  and  themselves  intersect  in  a  straight  line. 

In  general,  three  planes  separate  space  into  eight  parts, 
called  triedrals;  have  three  lines  of  intersection;  and 
have  one,  and  only  one,  common  point. 


~9^ 


(a)  Fig.  284.  (6) 

Fig.  (a)  represents  the  three  planes  so  placed  as  to  be 
perpendicular  to  each  other.     In  Fig.  (&)  they  are  oblique. 
Particular  cases  may  arise ;  for  instance : 

1.  The  third  plane  may  be  parallel  to  the  line  of  inter- 
section of  the  first  and  second.  The  three  lines  of  inter- 
section will  then  be  parallel. 

In  this  case  space  is  separated  into 
seven  parts. 

2.  The  third  plane  may  pass  through 
the  line  of  intersection  of  the  first  and 
second.  In  this  case  space  is  separated 
into  six  parts. 

206 


Fig.  285. 


206         ELEMENTS  OF  GEOMETRY. 

3.  The  three  planes  may  be  parallel  to  each  other.  In 
this  case  they  are  sometimes  said  to  intersect  at  infinity. 
They  separate  space  into  four  parts. 

When  three  planes  form  eight  triedrals,  the  common 
point  (/  in  the  figure)  is  called  the  ver- 
tex of  each. 

The  lines  of  intersection  of  the  planes 
form  the  edges  of  the  triedrals. 

IJ,  IK,  and  IH  are  edges. 

Fig    286 

The  angles  that  the  edges  make  with 
each  other  are  called  the  facial  angles. 

The  angles  that  the  planes  make  with  each  other  (as 
previously  noted)  are  called  diedrals. 

126.  Theorem.  If  the  vertex  of  a  triedral  be  taken  as 
the  centre  of  a  sphere  of  any  radius,  the  intersections  will 
form  a  sjjhei'ical  triangle,  the  sides  of  which  will  be  the 
measures  of  the  facial  angles ;  and  the  angles  of  which  will 
be  the  measures  of  the  diedrals. 

With  /  as  a  centre  and  ID  as  a  radius  describe  the 
sphere. 

The  DE  is  the  measure  of  the 

Z  DIE ;  the  EF  is  the  measure 

of  the  Z  EIF,  and  the  FD  is  the 

XT, 

measure  of  the  Z  FID.  _     .-_ 

If  at  any  of  the  vertices  of  the 
spherical  triangle,  as  F,  the  measure  of  the  diedral  be 
drawn  (§  106)  and  be  FO  and  FG,  they  will  (§  116) 
make  the  same  angle  as  do  the  arcs  FE  and  FD. 

Q.  E.  D. 


TRIEDRALS.  207 

Exercises.  —  1.   Show  that  if  two  facial  angles  of  a  triedral  are 
equal,  the  diedrals  opposite  them  will  be 
equal. 

Let  IJ,  IH,  and  IK  represent  the  edges 
of  the  triedral  *  of  which  Z  EID  =  Z  EIF. 

Conceive  a  sphere  with  /  as  a  centre 
and  any  radius,  say  IE,  intersecting  the 

planes  of  the  triedral  in  ED,  DF,  and  EF. 

Since  Z  EID  =  Z  EIF,  ED  =  EF.    .:  (§  124)  Z  EDF  =  Z  EFD. 
Hence  from  the  theorem,  the  diedrals  J-HI-K  and  J-KI-H  are 
equal. 

2.  Show  that  if  two  diedrals  of  a  triedral  are  equal,  the  facial 
angles  opposite  them  will  be  equal. 

3.  Show  that  in  any  triedral  the  sum  of  any  two  facial  angles 
will  be  greater  than  the  third. 

4.  Show  that  the  facial  angles  and  the  diedrals  of  a  triedral 
and  its  vertical  triedral  will  be  equal. 

5.  Show  that  the  sum  of  the  facial  angles  of  a  triedral  will  be 
between  0°  and  360°. 

6.  Show  that  the  sum  of  the  diedrals  of  a  triedral  will  be 
between  two  and  six  right  angles. 

7.  Show  that  two  triedrals  having  the  three  facial  angles  of 
one  equal  to  the  three  facial  angles  of  the  other,  and  similarly 
arranged,  are  superimposable. 

127.  Definition.  If  two  triedrals  can  be  so  placed  that 
the  edges  of  one  are  perpendicular  to  the  faces  of  the 
other,  they  are  said  to  be  supplementary. 

Let  the  two  triedrals  be  so  placed  that  their  vertices 
coincide ;  VK  being  perpendicular  to  the  face  A  VC,  VE 
perpendicular  to  the  face  AVB,  and  VD  perpendicular 
to  the  face  5  FO. 

*NoTE.  —  The  plane  MN  is  introduced  simply  for  the  purpose 
pf  enabling  the  beginner  to  more  readily  comprehend  the  figure. 


208 


ELEMENTS  OF   GEOMETRY. 


A  sphere  with  Fas  a  centre  would  be  intersected  by  the 
faces  so  as  to  form  on  its  surface  two  spherical  triangles. 

V 


Fig.  2S9. 

VK  passing  through  the  centre  of  the  sphere,  and 
being  perpendicular  to  the  plane  AVC  will  pierce  the 
surface  of  the  sphere  in  the  poles  of  the  arc  which  make 
the  intersection  of  the  sphere  with  the  plane  AVC. 

VE  will  pierce  the  surface  in  the  poles  of  HJ-,  and 
VD,  in  the  poles  of  JQ. 

These  poles  will  be  the  vertices  of  a  spherical  triangle,* 
polar  to  the  spherical  triangle  HJQ. 

We  have  seen  (§  122)  that  if  a  spherical  triangle  is 
polar  to  a  second  spherical'  triangle,  the  second  is  polar 
to  the  first. 

Hence :   VA  is  perpendicular  to  the  face  KVE, 

VB  is  perpendicular  to  the  face  DVE, 

and  VC  is  perpendicular  to  the  face  D  VK. 

The  relations  existing  between  the  parts  of  supple- 
mentary triedrals  are  therefore  mutual  ones. 

Applying  further  the  relations  established  in  §  122,  we 
have  the  following 

*  Note.  —  This  is  omitted  from  the  figure  in  order  to  save  con- 
fusion. 


POLYEDRALS. 


209 


Theorem.  The  facial  angles  of  each  of  two  siqjple- 
mentary  triedrals  are  the  supplements  of  the  opposite 
diedrals  of  the  other. 


128.    Problems.  —  1.   Show  that  the  sum  of  the  facial 
of  a  polyedral  angle  is  less  than  360°. 

Conceive  a  sphere  of  any  radius 
having  V  for  its  centre  and  inter- 
secting the  polyhedral. 

Then  apply  Ex.  11,  §  124. 

2.  Show  that  the  sum  of  the  die- 
drals will  be  between  the  limits, 
2n  —  4,  and  2n  rt.  ^i,  n  being  the 
number  of  faces.  i'Kj.  290. 


angles 


CHAPTER   XII. 

Surfaces. 

129.  Definitions.  1.  If  a  portion  of  space  be  enclosed 
by  planes,  the  portion  enclosed  is  called  a  polyedron. 

2.  As  has  been  seen,  three  planes,  in  general,  intersect 
in  a  point  and  do  not  enclose  any  portion  of  space.  It 
requires  a  fourth ;  although  four  planes  may  intersect  so 
as  not  to  enclose  a  portion  of  space. 

They  may  all  pass  through  one  line, 
they  may  be  parallel,  or  they  may  all 
pass  through  one  point. 

But  in  general  they  will  enclose  a 
portion  of  space,  and  the  portion  of  Fig.  291. 

space  is  called  a  tetraedron.  The  figure  V-ABC  is  a  tetra 
edron.  It  has  four  triangular  faces  and  four  vertices 
(points  through  which  three  planes  pass). 

Any  triangular  face  may  be  considered  the  base,  and 
the  vertex  not  in  the  plane  of  the  base  is  called  the  vertex 
opposite  the  base. 

'  3.  If  three  planes  intersect  in  parallel  lines 
and  the  figure  be  intersected  by  two  parallel 
planes,  the  enclosed  portion  of  space  is  called 
a  triangular  prism. 

4.   If  any  number   of  planes   intersect   in         '/    / 
parallel  lines,  and  these  parallel  lines  be  inter-      fig.  292. 

210 


A  PRISM. 


211 


Fig.  293. 


sected  by  two  parallel  planes,  the  enclosed  portion  of 
space  is  called  a  prism. 

The  prism  takes  its  name  from  the 
polygonal  intersections  of  the  parallel 
planes,  called  bases. 

In  the  above  figure  the  prism  is  hexag- 
onal. 

5.  The  parallel  lines  of  intersection,  as 
AB,  are  called  edges. 

6.  Plane  figures  situated  like   ABGD 
are  called  lateral  faces.     The  student  should  prove  that 
they  are  parallelograms. 

7.  If  the  edges  are  perpendicular  to  the  bases,  the 
prism  is  called  right;  if  not  perpendicular,  the  prism  is 
called  oblique. 

A  right  section  is  made  by  passing  a  plane  perpendicu- 
lar to  the  edges. 

8.  The  perpendicular  distance  between  the  parallel 
planes  forming  the  bases  is  called  the  altitude  of  the 
prism. 

9.  If  the  secant  planes  are  not  parallel,  the  enclosed 
portion  of  space  is  called  a  truncated  prism. 

10.  If  the  vertices  of  a  plane  polygon  be  joined  to  a 
point  without  the  plane  of  the  polygon,  the  figure  thus 
determined  is  called  a  pyramid. 

The  polygon  is  called  the  base,  and 
the  given  point,  P,  is  called  the  vertex. 

The  triangles  determined  by  the 
vertex  and  the  sides  of  the  polygon 
are  called  lateral  faces ;  and  the  sides  fig.  294. 

of  the  triangles  not  in  the  base  are  called  lateral  edges. 


212  ELEMENTS  OF  GEOMETRY. 

The  altitnde  is  the  perpendicular  distance  of  the  vertex 
from  the  base. 

11.  Pyramids  are  classified  in  several  ways : 

(a)  Convex,  if  the  base  is  a  convex  polygon ; 
Concave,  if  the  base  is  a  re-entrant  polygon. 

(b)  Triangular,  Quadrangular,  Pentagonal,  etc.,  depend- 
ing upon  the  number  of  angles  in  the  base. 

(c)  Regular,  if  the  base  is  a  regular  polygon  and  the 
projection  of  the  vertex  its  centre.  A  regular  pyramid 
is  frequently  called  a  right  pyramid. 

Irregular,  if  the  base  is  not  a  regu- 
lar polygon,  or  if  the  vertex  is  not 
projected  in  the  centre  of  the  base. 

12.  The  altitude  of  a  triangle  form-  yiq.  295. 
ing   one   of    the   faces   of   a  regular 

pyramid  is  called  a  slant  height  of  the  pyramid. 

130.  Theorem.  Sections  made  by  parallel  planes  inter- 
secting all  the  edges  of  a  prism  form  equal  polygons. 

Let  ABCDE  and  FQHJK  be  paral- 
lel i)lane  sections.     Then, 
AB  is  parallel  to  FG  (Prob.  1,  §  114), 
EC  is  parallel  to  GH  (Prob.  1,  §  114), 
Z  jiBC  =  ZFGH  (Prob.  9,  §  114). 
The  same  course  being  pursued  about 
the  perimeters  of  the  polygons,  they  are 
found  to  have  the  sides  and  angles  of 
one  equal  to  the  corresponding  parts  of  the  other.     They 
are  therefore  equal.  q.  e.  d. 


A  CYLINDRICAL  SURFACE. 


213 


Exercises.  —  1.    Show  that  if  a  plane  be  passed  parallel  to  the 
base  of  a  pyramid,  the  intersection  will  be  a 
polygon  similar  to  the  base. 

2.  Show  that  the  area  of  the  section  will  be 
to  the  area  of  the  base  as  the  squares  of  the 
distances  from  the  vertex.  

3.  Show  that  if  the  section  parallel  to  the 

base  bisect  the  altitude,  the  area  of  the  section  will  be  one-fourth 
the  area  of  the  base. 

4.  Show  that  the  lateral  area  of  a  prism 
equals  the  rectangle  of  the  perimeter  of  a  right 
section  and  a  lateral  edge. 


131.  Definitions.  1.  If  a  straight  line, 
not  in  tlie  plane  of  a  curve,  be  moved 
parallel  to  itself,  and  any  point  of  tlie 
line  move  along  the  curve,  a  cylindrical 
surface  will  be  generated. 


Fig.  298. 


The  straight  line  is  called  the  generatrix,  —  or  an  element 
of  the  surface,  —  and  the  curve  is  called  the  directrix. 

These  terms  are  interchange- 
able ;  for  the  curve  may  be  caused 
to  move  so  that  each  point  of  it 
will  generate  a  straight  line  par- 
allel to  a  given  straight  line  and 
will  generate  equal  lengths  in 
equal  intervals. 

Fig  299 

2.   Any  section  not  parallel  to 
the  straight  line  elements,  may  be  considered  as  the  base. 
The   character   of  the   base,  which   may  be  any  plane 
curve,  closed  or  not,  determines  the  general  character  of 
the  cylindrical  surface. 


214 


ELEMENTS   OF   GEOMETRY. 


Fig.  800. 


3.  A  cylinder  is  the  definite  portion  of  a  cylindrical 
surface  and  vokime,  included  between  two  parallel  bases. 

4.  If  the  elements  of  the  cylinder  are  perpendicular 
to  the  plane  of  the  base,  it  is  called  a 
right  cylinder, 

5.  The  cylinder  most  frequently  con- 
sidered is  a  right  cylinder  with  a  cir- 
cular base. 

6.  If  the  base  of  a  cylinder  be  a 
closed  curve,  and  if  a  polygon  be  in- 
scribed within  the  base,  and  the  elements  of  the  cylin- 
der through  the  vertices  of  this 
polygon  be  drawn,  an  inscribed 
prism  will  be  formed ;  the  vol- 
ume of  which  will  be  less  than 
the  volume  of  the  cylinder;  and 
the  area  of  the  bases  of  which 
will  be  less  than  the  area  of 
the  bases  of  the  cylinder. 

7.  If  under  the  same  con- 
ditions as  in  Def.  6  a  polygon  be  circumscribed  about 
the  base,  and  elements 
of  the  cylinder  be  drawn 
to  the  points  of  tangency, 
these  elements  and  the 
sides  of  the  polygon  will 
determine  planes  which 
are  said  to  be  tangent  to 
the  cylinder.  The  num- 
ber of  such  tangent  planes  ^'°-  ^^^2. 

will  equal  the  number  of  sides  of  the  polygon,  and  the 


Fig.  301. 


PRISMS. 


215 


figure  determined  by  them  will  be  a  prism  circumscrib- 
ing the  cylinder. 

This  prism  will  exceed  the  cylinder  in  volume,  and 
the  area  of  its  bases  will  be  greater  than  the  bases 
of  the  cylinder. 

132.  Definition.  A  parallelepiped  is  a  figure  formed  by 
six  planes,  parallel  two  and  two. 

The  lines  of  intersection  bound  the  six  faces. 

A  rectangular  parallelopiped  is  one  whose  angles  are 
all  right  angles. 

Exercises.  —  1.  Show  that  the  opposite  faces  of  a  pai-allelo- 
piped  are  equal  parallelograms. 


Fig.  303. 


Fig.  304. 


2.  Show  that  the  superficial  area  of  a  rectangular  parallelo- 
piped is  2  («&  +  ac  +  be). 

3.  Show  that  the  lateral  area  of  a  regular  pyramid  equals  half 
of  the  product  of  the  slant  height  by  the  perimeter  of  the  base. 


Fig.  305. 


Fig.  300. 


4.   Show  that  to  determine  the  lateral  area  of  an  oblique  pyra- 
mid the  area  of  each  triangle  will  have  to  be  determined  separately. 


216 


ELEMENTS  OF   GEOMETRY. 


5.  Show  that  the  cylinder  about  which  prisms  are  circum- 
scribed, or  within  which  prisms  are  in- 
scribed, will  be  the  limit  toward  which 
these  prisms  approach,  both  in  superficial 
area  and  in  volume,  as  the  number  of  lateral 
faces  of  the  prisms  is  made  to  increase. 


Fio.  SOT. 


133.  Definitions.  We  have  in  §  3 
described  a  curve  as  generated  by  a 
point  in  motion. 

If  a  point  move  on  a  curve  from 
one  position  to  another,  it  will  have 
moved  a  certain  distance. 

If  it  move  the  same  distance  on  a  straight  line,  the 
portion  of  the  straight  line  moved  over  is  said  to  be  the 
rectilinear  development  of  the  portion  of  the  curve. 

Illustration.  —  If  a  straight  line  be  tangent  to  a  circle 
at  T,  and  the  circle  then  be  rolled  upon  the  tangent  until 
the  point  T  again  comes  into  the  straight  line  at  (7^, 


T' 


Fig.  SOS. 


the  segment  of  the  straight  line  (TT^  over  which  the 
circle  has  rolled  is  the  rectilinear  development  of  the 
circumference. 

Theorem.     A  surface  that  can  be  generated  hy  a  straight 
line  moving  parallel  to  itself,  can  be  developed  in  a  plane. 


CYLINDERS.  217 

Let  HK  represent  a  cylinder,  the  elements  of  which 
are  oblique  to  the  parallel  bases. 

If  through  any  point,  P,  of  the  surface,  an  element  be 
drawn,  and  a  plane  be  passed  through  the  point  perpen- 
dicular to  the  element,  it  will  be  perpendicular  to  all  the 


Pre.  809. 

elements  of  the  cylinder,  will  intersect  the  cylinder  in 
a  curve,  and  will  be  a  right  section. 

If  at  P  a  tangent  to  the  right  section  be  drawn,  the 
plane  determined  by  this  tangent  and  the  element  will 
be  a  tangent  plane  to  the  cylinder. 

If  the  cylinder  be  rolled  on  the  plane,  the  right  section 
will  be  developed  into  a  straight  line  in  the  tangent 
plane  and  every  position  of  the  generating  element  of 
the  cylinder  will  come  into  the  tangent  plane,  —  being 
perpendicular  to  the  right  section  and  parallel  to  PT, 
after  development  as  well  as  before. 

When  the  original  element  of  tangency  shall  have 
returned  to  the  tangent  plane,  the  cylinder  will  have 
made  a  complete  roll,  and  every  point  and  element  of  its 
surface  will  have  come  in  contact  with  the  plane. 

The  cylinder  may  therefore  be  developed  on  a  plane. 


218 


ELEMENTS   OF   GEOMETRY. 


134.  Theorem.  Parallel  plane  sections  of  a  cylinder 
are  equal  Jigures. 

A  cylinder  being  gener- 
ated by  a  straight  line  mov- 
ing parallel  to  itself  and 
passing  through  every  point 
of  a  given  curve,  is  such  a 
surface  that,  being  cut  by 
parallel  planes,  the  elements 
included  between  them  will  all  be  of  equal  length. 

If  one  section  be  moved  parallel  to  itself,  and  so  that 
each  point  of  the  section  shall  follow  the  element  passing 
through  it,  it  will  fall  upon  the  other  section  and  coincide 
with  it  point  for  point.  Q.  E.  d. 

Let  the  cylinder  the  bases  of  which  are  B  and  B'  be  a 


Fig.  810. 


given  cylinder  with  oblique  bases,  and  MN  a  plane  tan- 
gent along  the  element  AD.  If  jB  be  a  right  section  and 
the  portion  between  R  and  B  be  moved  toward  B'  so  that 
each  point  shall  fall  on  the  element  passing  through  it, 


CYLINDERS.  219 

and  each  point  be  moved  a  distance  equal  to  the  portion  of 
the  elements  between  B  and  B',  B  will  come  to  coincide 
with  B',  and  R  will  have  advanced  to  the  position  R'. 

Thus  it  is  seen  that  the  convex  area  and  the  volume  of 
the  oblique  cylinder  are  equivalent  to  the  convex  surface 
and  the  volume  of  the  right  cylinder  RR'. 

If  this  right  cylinder  be  rolled  upon  the  tangent  plane 
until  the  element  of  initial  contact  shall  again  return  to 
the  plane,  the  surface  of  the  cylinder  will  be  developed 
on  the  tangent  plane  in  the  form  of  a  rectangle,  one  side 
of  which  is  the  circumference  of  a  right  section,  and 
the  adjacent  side  is  the  portion  of  an  element  included 
between  the  bases. 

Hence  the  following 

Theorem.  The  area  of  the  convex  surface  of  a  cylinder 
betiueen  parallel  plane  sections  equals  a  rectangle,  one  side 
of  lohich  is  the  perimeter  of  a  right  section,  and  the  adjacent 
sides  the  segment  of  an  element  included  between  the  p)lane 
sections. 

Exercises. — 1.  Find  the  convex  (i.e.  curved)  surface  of  a 
right  cylinder,  the  diameter  of  the  base  being  2  feet,  and  the  alti- 
tude 100  feet. 

2.  How  many  square  feet  of  sheet  iron  will  be  required  to  make 
300  feet  of  10-inch  pipe,  allowing  one-sixth  of  the  entire  amount 
of  sheet  iron,  for  lapping  and  waste  ? 

135.  Definitions.  If  a  straight  line  pass  through  a 
fixed  point  and  follow  a  curve,  it  will  in  general  describe 
a  conical  surface.  The  only  exception  is  when  the  curve 
is  a  plane  one  and  the  fixed  point  is  in  the  plane  of  the 
curve. 

The  point  is  called  the  vertex  of  the  conical  surface, 


220 


ELEMENTS   OP  GEOMETRY. 


Fig.  312. 


the  curve,  through  the  different  points  of  which  the 
straight  line  passes,  is  called  the  directrix,  and  the  straight 
line  is  called  an  element  of  the  surface.  The  parts  of 
the  cone  separated  from  each  other  by  the  vertex  are 
called  nappes. 

If  the  directrix  be  a  circle  and  the  vertex  be  in  a 
perpendicular  to  the  plane  of 
the  circle  through  its  centre, 
the  surface  generated  is  called 
a  right  circular  cone,*  and  the 
perpendicular  is  called  the 
axis  of  the  cone. 

In  all  cones  the  surface  is 
not  limited  in  extent,  but  for 
certain  purposes  limited  por- 
tions may  be  considered. 

Any  plane  section  may  be  taken  as  the  base. 

A  frustum  is  the  part  of  one  nappe  that  lies  between 
parallel  planes. 

If  a  straight  line  be  drawn  tangent  to  the  directrix  of 
a  conical  surface,  the  plane  determined  by  that  tangent 
and  an  element  will  not  con- 
tain the  adjacent  elements, 
and  is  said  to  be  tangent  to 
the  conical  surface. 

A  tangent  line  to  a  curve 
may  be  rolled  along  on  the 
curve.     In  each  position  it  will  determine  with  the  ele- 

*NoTE.  — The  right  circular  cone,  when  cut  by  planes  in  differ- 
ent positions,  gives  rise  to  the  plane  curves  called  conic  sections, 
the  most  noted  and  most  studied  of  all  curves. 


Fig.  818. 


THE    CONE. 


221 


ment  through  the  point  of  contact,  a  tangent  plane.  So 
that  a  tangent  plane  to  a  conical  surface  may  be  rolled 
upon  the  surface  from  any  position  to  any  other  position. 

Instead  of  rolling  the  plane  on  the  surface,  the  sur- 
face may  be  rolled  on  the  plane,  and  thus  we  see  that 
any  conical  surface  may  be  developed  on  a  tangent 
plane ;  the  directrix  in  general  falling  in  a  curve,  and 
every  point  of  the  surface  coming  into  the  plane  on 
which  the  development  is  made. 

All  tangent  planes  to  a  cone  will  pass  through  the  ver- 
tex, and  if  one  najJjoe  is  entirely  on  one  side  of  the  tan- 
gent plane,  the  other  nappe  will  be  on  the  other  side. 

Exercise.  —  Show  that  all  plane  sections  parallel  to  the  base  of 
a  right  circular  cone  will  be  circles,  the  circumferences  of  which 
are  to  each  other  as  their  distances  from  tlie  vertex  ;  and  the  areas 
of  which  are  to  each  other  as  the  squares  of  these  distances. 

136.  If  at  a  point  T  of  the  base  of  a  right  -circular 
cone,  a  tangent  to  the  circle  be  drawn,  it  will,  with  the 
element  VT,  determine  a  tangent  plane  MN. 


Fig.  314. 


If  the  cone  be  rolled  on  the  tangent  plane  until  the 
element  VT  again  comes  into  the  plane,  the  portion  of 


222 


ELEMENTS   OF   GEOMETRY. 


the  surface  between  the  base  and  the  vertex  will  be 
developed  into  the  sector  of  a  circle,  having  the  slant 
height  of  the  cone  for  its  radius  and  an  arc  equal  to  the 
circumference  of  the  base. 

Since  by  §  103,  Ex.  1,  we  have  the  area  of  a  sector  equal 
to  half  the  product  of  the  arc  and  the  radius,  we  have  the 

Theorem.  The  convex  area  of  a  right  circular  cone 
equals  half  the  product  of  the  slant  height  by  the  circum- 
ference of  its  base. 

Definitions.  1.  If  a  polygon  be  circumscribed  about 
the  base  of  a  cone  and  its  vertices  be  joined  with  the  ver- 
tex of  the  cone,  a  circumscribed  pyra- 
mid will  be  formed. 

2.  If  the  vertices  of  an  inscribed 
polygon  be  joined  with  the  vertex  of 
the  cone,  an  inscribed  pyramid  will  be 
formed. 

Fig.  315. 

Exercises.  —  1.   Show  that  the  surface 
and  volume  of  a  cone  are  the  limits  toward  which  the  surfaces  and 
volumes  of  circumscribed  and  inscribed  pyramids  approach  as  the 
number  of  sides  is  increased. 

2.  Show  the  same  for  the  volume  and  the 
lateral  surface  of  a  frustum. 

Note. — A  cone  not  right  is  called 
oblique ;  and  while  its  surface  may  be 
developed  on  a  tangent  plane,  it  will  not 
be  developed  into  a  sector  of  a  circle,  and 
the  arc  cannot  in  general  be  determined  by  the  methods  of  ele- 
mentary geometry. 

137.  Problems.  —  1.  Find  an  algebraic  expression  for  the 
convex  surface  of  a  frustum  of  a  right  circular  cone,  the  planes 
being  passed  perpendicular  to  the  axis. 


Fio.  316. 


THE   CONE. 


223 


:  slant  height  Kn, 


Let  C  represent  the  circumference  of  the  larger  section  and  c 
represent  the  circumference  of  the 
smaller  section. 

'  '  \q 

q  =  slant  height  HV, 

L  =  lateral  surface  of  VC, 

I  =  lateral  surface  of  Vc, 
F  =  lateral  surface  of  frustum  Cc, 


F=L-l. 


Fig.  317. 


I  =  \  rq, 


(§  136) 
(§  136) 


F=\C{s  +  q)-\cq 
=  \Cq-\  eg  +  I  Cs 
=  \{C-c)q  +  \  Cs, 

a  form  of  expression  for  the  required  area,  but  one  that  involves 
q,  a  Une  which  is  not  a  part  of  the  frustum. 

In  order  to  eliminate  it  from  the  expression,  it  is  necessary  to 
find  relations  betvreen  it  and  lines  of  the  frustum. 

C^s  +  q 
c         q    ' 
Cq  =  cs-i-  cq, 
Cq  —  cq  =  cs, 
(C—  c)  q  =  cs. 

Substituting  this  value  of  {C  —  c)  q  in  the  last  obtained  expres- 
sion for  F,  we  have, 

F=h  cs  +  ^  Cs  or  F=^  (C+c)  s. 
:.  F  =  I  (2  TT  B  +  2  TT  r)  s  -  TT  (r  +  E)  s. 

2.  Show  that  ^  ic+  C)  equals  the  circumference  of  a  middle 
section,  and  that  the  area  of  the  frustum  may  be  described  as 
the  circumference  of  the  middle  section  by  the  slant  height: 
'r±_B\^ 
2 


224 


ELEMENTS   OF   GEOMETRY. 


M, 


3.   Find  an  algebi'aic  expression  (formula)  for  the  frustum  of  a 
cone  in  terms  of  the  altitude  EG. 

From  Prob.  2  we  have : 

F=2irJM-BD.  D 

The  analysis  suggests  the  drawing  of 
DNW  to  GJS  and  MQ±  to  BD ;  from 
which, 

JM_IW 
'QM     BD 

or  JM  BD=  QM-  ND, 

..  F=2irQM'ND  =  2irQM-EG.  q.  e.  f. 


Q 
U 

Fio.  318. 


The  expression  last  deduced  shows  that  the  area  generated  by 
revolving  a  segment  of  a  straight  line 
about  another  straight  line  in  the  same 
plane,  will  generate  a  surface  which 
would  equal  the  convex  area  of  a  cylin- 
der having  for  its  radius  the  perpen- 
dicular from  the  middle  point  of  the 
generating  line  to  the  axis  and  for  its 
altitude  the  projection  of  the  given  seg- 
ment on  the  axis. 

Pic    319 

4,  Show  that  if  a  regular  polygon  of 
an  even  number  of  sides  be  revolved  about  a  diameter  of  the  cir- 
cumscribed circle,  which  is  also  a  diago- 
nal of  the  polygon,  as  an  axis,  the  area 
generated  will  equal  the  convex  surface  f 
of  a  right  cylinder  having  for  the  radius 
of  its  base  the  apothem  of  the  polygon, 
and  the  diameter  of  the  circumscribed 
circle  for  its  altitude. 


E 


Fio.  820. 


Or  expressed  differently:  The  area  generated  by  the  rotation  of 
a  regular  polygon,  as  above  described,  will  equal  the  convex  area 
of  a  right  cylinder,  having  for  the  diameter  of  the  base  the  diam- 


SPHERICAL   SURFACE. 


225 


eter  of  the  inscribed  circle,  and  for  its  altitude  the  diameter  of  the 
circumscribed  circle. 


M 


Fm.  321. 


138.  The  surface  of  a  cone  or  of  a  frustum  of  a  cone 
may  be  considered  as  generated  by  a  circle,  the  plane 
of  which  moves  parallel  to  itself  and  the  radius  of  which 
varies  as  its  distance  from  a  given  point.  For  if  we  have 
a  right  circular  cone  or  frustum,  sections  perpendicular 
to  the  axis  will  produce  circles,  the  radii  of  which  will 
be  to  each  other  as  the  distances  of  the  sections  from 
the  vertex. 


Fig 


The  surface  of  a  sphere  or  of  a  zo7ie  (§  116)  may  be 
generated  by  moving  a  circle  parallel  to  itself  from  A 


226 


ELE^rENTS   OF   GEOMETRY. 


toward  X,  remaining  perpendicular  to  AX,  and  so  chang- 
ing the  radius  that  its  square  shall  equal  the  product  of 
AQ  and  QX. 

For  if  we  have  a  sphere,  the  radius  of  any  plane 
section  will  be  a  mean  proportional  between  the  seg- 
ments into  which  the  diameter,  perpendicular  to  the 
section,  is  separated  by  it. 

Theorem.  If  a  chord  of  a  given  circle  be  a  tangent  to 
a  concentric  circle,  and  the  extremities  of  the  chord  be 
joined  to  tJie  centre,  and  the  figure  thus  determined  be 
revolved  about  a  non-intersecting  diameter,  the  surface  gen- 
erated by  the  exterior  arc  will  be  greater  than  the  surface 
generated  by  the  chord,  which  will  be  greater  than  the  surface 
generated  by  the  interior  intercepted  arc. 

By  §  103,  BD>BD>KE. 

If  any  radius  CP  be  drawn  intersecting  the  inner 
circle  at  J,  and  the  chord  at  T,  P 
will  be  at  a  greater  distance  from 
the  axis  than  T,  and  T  will  be  at  a 
greater  distance  than  J.  So  that 
the  circumference  generated  by  P 
will  be  greater  than  that  generated 
by  T',  which  in  turn  will  be  greater 
than  the  circumference  generated 
byj:  ^ 

Eevolved  about  AX,  BD  will  gen- 
erate a  zone,  BD  the  frustum  of  a  cone,  and  KE  a  zone. 

These  may  also  be  generated,  as  seen  in  the  early  part 
of  this  article,  by  circumferences,  the  planes  of  which  are 
perpendicular  to  the  axis. 


SPHERICAL   SURFACE.  227 

In  the  cases  now  under  consideration,  the  larger  cir- 
cumferences will  be  moved  the  greater  distance  and  so 
will  generate  the  larger  area. 

The  zone  formed  by  the  revolution  of  BD  will  be 
greater  than  the  frustum  formed  by  the  revolution  of  BD, 
and  that  in  turn  will  be  greater  than  the  zone  formed  by 
the  revolution  of  KE. 

Hence  the  theorem  is  established. 

139.  Problem.  To  find  the  expression  for  the  surface 
of  a  sphere  in  terms  of  the  square  on  its  radius. 

As  a  consequence  of  the  last  article,  if  a  regular 
polygon  of  an  even  number  of  sides,  together  with  its 
circumscribed  and  inscribed  circumferences,  be  revolved 
about  a  diagonal  passing  through  a  pair  of  opposite 
vertices,  the  surface  generated  by  the  polygon  will  be 
less  than  the  surface  of  the  circumscribed  sphere  and 
greater  than  the  surface  of  the  inscribed  sphere. 

In  Prob.  4,  §  137,  it  is  shown  that  the  area  generated 
by  the  polygon  equals  the  convex  surface  of  a  right 
cylinder,  having  for  the  diameter  of  its  base  the  diameter 
of  the  inscribed  circle,  and  for  its  altitude  the  diameter 
of  the  circumscribed  circle. 

If  we  should  retain  the  same  circumscribing  circle 
and  should  double  the  number  of  sides  of  the  polygon, 
the  diameter  of  the  inscribed  circle  would  be  increased, 
and  the  area  generated  by  the  polygon  of  the  increased 
luimber  of  sides  would  be  equivalent  to  a  cylinder  having 
the  same  altitude  as  before,  but  having  a  greater  diameter 
of  base. 

Each  time  the  number  of  sides  of  the  polygon  is 
doubled,  the  diameter  of  the  cylinder  having  the  equiva- 


228         ELEMENTS  OF  GEOMETRY. 

lent  convex  area  will  be  increased,  but  its  altitude  will 
not  be  changed. 

We  may  conceive  of  this  operation  being  performed  as 
many  times  as  we  please,  and  at  each  one  the  surface 
generated  by  the  polygon  will  be  equivalent  to  the  con- 
vex surface  of  a  cylinder  that  remains  constant  in  altitude, 
but  the  diameter  of  the  base  of  which  is  increased. 

This  increase  of  diameter  is  limited,  because  the  di- 
ameter of  the  inscribed  circle  will  always  be  less  than  the 
diameter  of  the  circumscribed  circle.  It  will,  however, 
approach  the  diameter  of  the  circumscribed  circle,  as  we 
conceive  the  number  of  sides  of  the  polygon  to  be  in- 
creased, and  the  difference  between  the  two  diameters 
may,  by  increasing  the  number  of  sides  of  the  polygon,  be 
reduced  to  as  small  a  quantity  as  we  please,  and  the  two 
spherical  surfaces  be  made  to  approach  as  near  as  we  please. 

As  we  perform  these  operations,  or  conceive  them  as 
being  performed,  there  are  two  definite  limits  toward  which 
the  area  generated  by  the  polygon  approaches  : 

These  are  the  surface  of  the  circumscribing  sphere,  and 
the  convex  surface  of  a  cylinder  having  both  diameter  of 
base  and  altitude  equal  to  the  diameter  of  the  circum- 
scribing sphere. 

Therefore,  by  the  limits  axiom,  the  surface  of  the  cir- 
cumscribing sphere  will  equal  the  con- 
vex surface  of  the  cylinder  having  for 
its  diameter  and  altitude  the  diame- 
ter of  the  circumscribing  sphere. 

The  convex  surface  of  the  cylinder 
is:  2wE-2R  =  AirJi^ 

.:  Spherical  surface  =  4  TriJ^ 


SPHERICAL   SURFACE.  229 

Note.  —  The  surface  of  the  sphere,  the  centre  of  which  is  0, 
equals  the  convex  surface  of  the  circumscribed  cylinder ;  and  equals 
four  times  the  area  of  the  base  of  the  cylinder  or  the  area  of  four 
great  circles.  Neither  the  convex  surface  of  the  cylinder  nor  the 
plane  surface  of  four  great  circles  can  be  applied  to  the  surface 
of  the  sphere  although  they  are  equal. 

Exercises.  —  1.  Show  that  the  surfaces  of  any  two  spheres  are 
to  each  other  as  the  squares  of  their  radii,  or  as  the  squares  of  any 
corresponding  lines. 

2.  Find  the  surface  of  a  sphere,  the  diameter  of  which  is  one 
and  a  half  feet. 

140.   Problem.     To  find  the  area  of  a  zone. 

The  arc  BD,  revolving  about  AK,  will  generate  a  zone. 

The  chord  BD  would  generate  the  convex  surface  of  a 
frustum,  the  area  of  which,  as  we  have 
seen  in  Prob.  3,  §  137,  would  equal  the 
convex  surface  of  a  cylinder,  having  the 
apothem  of  BD  for  the  radius  of  its 
base,  and  the  projection  QK  for  its 
altitude. 

If  the  arc  BD  be  bisected  and  the 
chords  BM  and  MD  be  drawn  and  then 
revolved  about  AX,  the  surface  will  be  the  sum  of  two 
frustums,  the  combined  convex  area  of  which  will  be  the 
convex  surface  of  a  cylinder  having  for  the  radius  of  its 
base  the  apothem  of  the  new  polygon,  and  for  its  altitude 
QK  (a). 

And  further  pursuing  the  matter  in  the  same  way  as 
in  the  preceding  article,  we  find  the  area  of  a  zone 
equals  the  convex  surface  of  a  cylinder  having  the  radius 
of  the  sphere  for  the  radius  of  its  base  and  the  altitude 
of  the  zone  for  its  altitude,  or  Z=2Trli-a. 


230  ELEMENTS   OP   GEOMETRY. 

141.  Definitions.  A  lune  is  the  portion  of  the  surface  of 
a   sphere   included  between  two  semi-  t) 

circles  (§  117).  /^'^  TN>^ 

Two  great  circles  of  a  sphere  separate  /  _j-4-l    \ 
its  surface  into  four  lunes.  c--^ — ^-\ — Ab 

The  angle  of  a  lune  is  the  angle  that  V    T  !   \aJ 
the  planes  of  the  great  circles  make  with      \.lij^Z^ 
each  other.     As   we  have  seen  (§  116), 
the  angle  between  the  planes  is  the  same  '°' 

as  the  angle  between  the  semicircles  at  their  intersection, 
and  that  is  measured  by  the  arc  of  a  great  circle  included 
between  the  arcs  forming  the  lune,  and  90°  from  the 
vertex. 

The  surface  of  a  sphere  may  be  considered  as  generated 
by  the  rotation  of  a  semicircumference  about  its  diameter 
as  an  axis. 

Starting  from  its  position  DAE,  the  semicircumference 
may  be  conceived  as  rotating  with  uniform  velocity  until 
it  shall  have  returned  to  its  initial  position.  During  this 
rotation  the  line  CA  rotates  with  uniform  velocity  and 
generates  the  angle  which  is  the  measure  of  the  diedral. 

Each  starts  at  the  same  time,  the  motion  is  uniform, 
and  the  motion  is  completed  at  the  same  time. 

When  the  semicircumference  shall  have  arrived  at  the 
position  DBE,  the  lune  A-DE-B  will  have  been  gen- 
erated and  it  will  be  the  same  fractional  part  of  the  entire 
surface  of  the  sphere  that  the  angle  ACB  is  of  300°. 
Lune  _  Angle  of  lune 
''■  Sphere"         360°         ' 

T  Angle  of  lune    q  i 

or  Lune  =  — ^ .  Sphere. 

360°  ^ 


SPHERICAL   SURFACE.  "  231 

Corollary.  Two  lunes  are  to  each  other  as  the  angles 
of  the  limes. 

L^_e_      I  ^    <^ 
S~360°'   S  ~360=" 

•  t^  =  l 
"   I      <{>' 

Exercise.  —  Show  that  the  sum  of  a  number  of  lunes  equals  a 
single  lune,  having  for  its  angle  the  sum  of  the  angles  of  the 
several  lunes. 

Beviark.  —  A  hemisphere  may  be  regarded  as  a  lune  having  an 
angle  of  180°,  and  a  sphere  as  a  lune  of  360°. 

142.  Problem.  Deduce  a  formula  *  for  the  area  of  a 
spherical  triangle. 

Let  ABC  represent  any  spherical  triangle,  one  side, 

B 


as  a  matter  of  convenience,  being  in  the  base  of  the 
hemisphere  on  which  the  spherical  triangle  is.  (See 
note,  §  121.) 

The  sp.  A  ABC  4-  sp.  A  BCD  =  Lune  A% 
sp.  AABC  +  sp.  A  ACE  =  Lune  £°, 
sp.  A  ABC  +s^.  ADEC=  Lune  G°,      p  jc  (^ 

3  sp.  A  ABC  +  sp.  A  BCD  +  sp.  A  ACE  +  sp.  ^^aEi  = 
Lune  A°  +  Lune  B^  +  Lune  C°, 

*  A  formula  is  the  algebraic  expression  of  a  law. 


232         ELEMENTS  OF  GEOMETRY. 

or  2  sp.  A  ABC  +  Hemi.  =  Lune  (A° -\- B"  +  C°), 
2  sp.  A  ABC  =  Lune  {A°  +  B°  +  C°)  -  Hemi., 
2  sp.  A  jLBC  =  Liine  (^^  +  ^  +  C°  -  180°), 

or         sp.  A  ABC  =  ?^  (A° -\- B^  +  C°  -  180°). 

Exercises.  —  1.  Deduce  a  formula  for  the  area  of  a  trirec- 
tangular  spherical  triangle. 

2.  Deduce  an  expression  for  sp.  A  ABC  in  terms  of  T  (a  tri- 
rectangular  spherical  triangle). 

3.  Deduce  a  formula  for  the  area  of  a  spherical  triangle  in 
terms  of  the  sphere. 

4.  Deduce  a  formula  for  finding  the  area  of  any  spherical  poly- 
gon, the  angles  of  which  are  given. 

Note.  —  It  has  been  shown  that  the  sum  of  the  angles  of  a 
spherical  triangle  is  greater  than  180°.  The  amount,  in  degrees, 
by  which  the  sum  of  the  angles  exceeds  180°,  is  called  the  spher- 
ical excess  {A°  +  B°  +  C°  -  180°). 

143.  Theorem.  Similar  surfaces  are  to  each  other  as 
the  squares  of  any  homologous  {corresponding)  lines. 

Let  P  and  p  represent  two 
similar  polyedrons. 

Analysis.  —  If  the  surfaces  of 
these  polyedrons  are  to  each 
other  as  the  square  of  any  horn-  ^^ 
ologous  lines,  any  correspond- 
ing portions  of  the  surfaces 
would  sustain  the  same  relation. 

Demonstration.  —  The  polyedra  being  similar,  are 
bounded  by  similar  plane  polygons,  similarly  placed. 
Thus  A  ABC  and  abc  are  similar  and  are  placed  with 


SIMILAR   SUKFACES.  233 

respect  to  the  similar  A  BCD  and  hcd  so  as  to  make  tlie 
equal  diedrals  A-BC-D  and  a-bc-d. 

The  same  may  be  said  of  any  other  adjacent  faces. 
By  §  78  we  have, 

A  ABC  _  A&_  BC^_  AG^ 
~Aabc'  ~W~W~  ^' 

A  BCD  _  BG'_  BD'_  CD" 
Abed  ~bc~  hdt~'^' 

We  have  the  common  ratio  ^=^.   so  all  the  ratios  are 

6c' 
equal,  and  by  §  79  we  have, 

A  ABC  +  A  BCD  _  B^_  :^_  B&  ^^ 
Aabc  +  Abcd     ~bc~ab'~  u"^'"' 

,      .  ABDE     BE^    BD^    . 

^g^^^'  -Abd^  =  ^  =  W'"'^' 

By  §  79  we  have, 

A  ABC  +  A  BCD  +  ABDE     BD''    Ib"   , 

■ =  —^~  = ,  etc. 

A  a6c  +  A  bed  +  A  bde  i)(f      ^j^'' 

In  the  same  manner  we  may  consider  each  of  the  cor- 
responding polygons  forming  faces  of  the  similar  poly- 
edra;  and  find  that  the  ratios  between  the  several 
similar  polygons  will  be  the  same  as  the  ratio  of  the 
squares  of  any  homologous  lines ;  and  the  sums  of  all  of 
them  will  have  the  same  ratio. 

There  are,  as  already  seen,  surfaces  entirely  curved,  as 
a  sphere,  and  surfaces  partly  curved  and  partly  plane. 

In  any  case  the  similar  surfaces  may  be  considered 
as  the  limits  toward  which  similar  inscribed  or  circum- 


234  ELEMENTS   OF   GEOMETRY. 

scribed  polyedra  approach  as  the  number  of  faces  is 
increased. 

Lines  may  be  chosen  that  shall  be  corresponding  lines 
in  the  two  similar  surfaces  under  consideration  and 
which  shall  be  corresponding  lines  in  the  approaching 
polyedra. 

The  surfaces  of  the  approaching  similar  polyedra  are 
to  each  other  as  the  squares  of  the  homologous  lines 
which  may  be  so  chosen  as  to  remain  unaltered  during 
the  approach. 

s,~  r 

^3-^'  etc 

These  surfaces  represented  by  Si,  s^,  S^,  s^,  S^,  Sg,  etc., 
which  are  similar  at  each  step  and  continually  approach 
the  curved  or  mixed  surfaces,  always  have  the  same  ratio 

— .     The  limits  toward  which  they  approach  will  have 

the  same  ratio  (§  96).     Hence  the  theorem. 

Note.  —  In  the  above  article  is  established  the  second  of  the 
three  great  principles  of  geometry  as  stated  in  §  69,  Note. 

Exercises.  —  1.  From  the  formula  for  the  surface  of  a  sphere, 
show  that  the  surfaces  of  two  spheres  are  to  each  other  as  the 
squares  of  their  radii. 

2.  Show  that  the  ratio  of  the  areas  of  similar  zones  on  different 
spheres  equals  the  ratio  of  the  squares  on  the  diameters  of  the 
spheres. 


SIMILAR    SUBFACES.  235 

3.  Making  use  of  a  spherical  blackboard,  show  how  to  find 
approximately  the  distance  from  New  York  to  Queenstown. 

4.  With  the  same  materials  find  approximately  the  distance 
from  San  Francisco  to  Yokohama,  and  show  how  near  to  the 
Aleutian  Islands  the  shortest  arc  will  pass. 


CHAPTER   XIII. 

Volumes. 

144.  Definitions.     1.   As  has  already  been  stated,  a  vol- 
ume is  an  enclosed  and  limited  portion  of  space. 
The  figure  may  be  real  or  imaginary. 

2.   In  §57  the  area  of  a  rectangle  is  represented  by 
ab,  the  product  of  two   ad- 
jacent sides. 

If  that  rectangle  be  moved 
perpendicularly  a  distance  c, 
the  volume  generated  is  rep- 
resented by  abc.  Expressed 
as  a  formula,  it  is,  V=  abc. 

The  figure  is  called  a  rectangular  parallelopiped.  It 
is  also  a  right  prism  with  rectangular  base. 

If  the  volume  of  one  rectangular  parallelopiped  be  rep- 
resented by  V,  and  the  volume  of  another  by  v,  the  three 
edges  meeting  at  a  vertex  in  the  one  being  a,  b,  and  c, 
and  in  the  other,  x,  y,  and  z,  we  have, 

V  =  abc,  V  =  xyz. 

By  division,  —  = — j    .■.  the 

V  xyz 

Theorem.  The  volumes  of  tioo  rectangular  parallelo- 
pipeds  are  to  each  other  as  the  products  of  the  three  edges 
meeting  at  a  vertex. 

236 


EQUIVALENT    VOLUMES.  237 

Esercises.  —  1.    Show  that  the  theorem  might  be  stated  thus : 
The  volumes  of  two  rectangular  parallelopipeds  are  to  each 
other  as  the  products  of  their  bases  and  altitudes. 

2.  If  the  bases  happen  to  be  equal,  they  are  to  each  other  as  their 
altitudes. 

3.  If  their  altitudes  happen  to  be  equal,  they  are  to  each  other 
as  their  bases. 

4.  Show  that  if  the  figures  are  similar,  their  volumes  will  be  to 
each  other  as  the  cubes  on  any  of  the  corresponding  lines. 

145.  Theorem.  An  oblique  prism  is  equivalent  to  a 
right  prism,  the  base  of  which  is  a  right  section  of  the 
oblique  prism,  and  the  altitude  of  which  is  equal  to  an  edge 
of  the  oblique  prism. 

Let  A-I  represent  the  oblique  prism,  and  K-N  a  right 
section,  and  QS  another  right  section  at  a  distance  from 
K-N  equal  to  an  edge,  AF,  of 
the  oblique  prism. 

Analysis.  —  If  the  right  prism 
K-S  be  equivalent  to  the  ob- 
lique prism  A-I,  the  truncated 
prism  K-I  being  common  to 
both,  the  truncated  prism  FS 
must  be  equivalent  to  the  trun- 
cated prism  A-]^. 

Demonstration. — Conceive  the  ^'**-  ^^• 

truncated  prism  A-N  as  moved  in  the  direction  of  the 
edges.  When  it  shall  have  been  moved  the  distance 
AF,  the  oblique  polygon  ABCDE,  forming  its  lower 
base,  will  coincide  with  FGHIJ,  forming  its  upper  base, 
the  right  section  KMNOP  will  coincide  with  the  right 
section  QRSTU,  and  the  lateral  faces  will  coincide,  so 


238 


ELEMENTS   OF   GEOMETRY. 


that  the  oblique  prism  A-I  will  be  converted,  without 
change  of  volume,  into  the  right  prism  K-S.  q.  e.  d. 

146.  Theorem.  An  oblique  parallelopiped  is  equivalent 
to  a  right  parallelopiped  having  an  equivalent  base  and  the 
same  altitude. 

Let  ABCDEFGH  represent  the  oblique  parallelopiped. 
The  figure  will  be  a  prism,  no  matter  which  of  the  six 
faces  be  taken  as  a  base. 

If  a  plane  section  be  made  through  ^  X  to  AD  and  if 


Gl 


E  M 


Fig.  332. 


a  parallel  plane  be 
passed  through  D,  we 
shall  have  (by  §  145) 
the  oblique  prism  con- 
verted into  an  equiv- 
alent right  prism, 
AIJKDONM. 

If  ADOK  be  considered  as  the  base  of  the  second 
prism,  the  edges  parallel  to  AI  will  be  oblique  to  it.  A 
plane  passed  through  AD  and  perpendicular  to  AI  will 
make  a  right  section,  APQD.  Another  right  section, 
IMRS,  may  be  passed  through  IM,  and  we  shall  have 
(by  §  145)  the  prism  A-N,  that  is  oblique  with  respect 
to  the  face  A-0,  converted  into  a  right  prism  A-R,  every 
angle  of  which  is  a  right  angle. 

In  these  changes  from  one  figure  to  an  equivalent  one, 
the  altitude  has  remained  the  same,  being  the  perpen- 
dicular distance  between  the  planes  H-C  and  G-D.  By 
the  first  change  the  base  H-C  was  converted  into  the 
rectangular  base  J-0,  equivalent  to  H-G.  And  by  the 
second  change  the  rectangular  base  J-0  was  converted 
into  the  equivalent  rectangular  base  S-Q. 


VOLUME   OP   A  PRISM. 


239 


These  changes  have  been  effected  without  change  of 
vohime.  q.  e.  d. 

Exercise.  —  Show  that  the  volume  of  any  oblique  parallelopiped 
is  represented  by  the  product  of  its  base  and  altitude. 

147.   Problem.     To  find  an  expression  for  the  volume 
of  a  triangular  prism. 

Let  ABCDEO  represent  an  oblique  triangular  prism. 

Analysis.  —  From   what  has  preceded  with  regard  to 
volumes,  it  seems  probable  that  the  volume  of  the  oblique 


Fig.  383. 


Fig.  334. 


triangular  prism  will  be  represented  by  the  product  of  its 
base  and  altitude. 

If  it  be  so,  we  must  arrive  at  the  determination  through 
the  parallelopiped. 

If  BH II  to  AC, 

CH II  to  AB, 

EK  W  to  DG, 

GK  W  to  DE, 

and  HK  be  drawn,  an  oblique  triangular  prism  will  be 

annexed  to  the  original  triangular  prism  so  as  to  form  a 

parallelopiped,  the  volume  of  which  is  represented  by  the 

product  of  base  and  altitude. 

The  two  triangular  prisms  have  equivalent  bases  and 


240         ELEMENTS  OF  GEOMETKY. 

the  same  altitude,  and  each  base  is  half  the  base  of  the 
parallelopiped. 

The  added  prism  we  suspect  is  equivalent  to  the  original 
prism,  but  it  cannot  be  substituted  for  the  original  prism 
and  be  made  to  occupy  the  same  space. 

If  we  attempt  the  substitution,  we  shall  have  the  base 
BHC,  coinciding  with  the  base  CAB,  by  being  reversed, 
but  the  elements  will  not  take  the  same  direction. 

We  may,  however,  convert  the  two  oblique  trianglar 
prisms  into  equivalent  right  prisms  having  for  their  bases 
right  sections,  MNP  and  NOP,  of  the  oblique  prisms,  and 
for  their  edges  (altitudes)  the  edges  of  the  oblique  prisms. 
These  right  prisms  may  be  substituted,  the  one  for  the 
other,  and  occupy  the  same  space.  But  they  are  separ 
rately  equivalent  to  the  oblique  triangular  prism 

Hence  the  oblique  prisms  are  equivalent. 

Parallelopiped      A-K  =b  -h. 


product  of  base  and 


Esercises. — 1.  Show  that  the  volume  of  any  prism  may  be 
represented  by  the  product  of  its  base  and  altitude. 

2.  Show  that  the  volume  of  the  prism  A-G  may  be  represented 
by:  (A MNP) (AD). 

8.  Show  by  the  method  of  limits  that  the  volume  of  any  closed 
cylinder  may  be  represented  by  the  product  of  base  and  altitude. 

4.  Show  that  the  diagonals  of  ^a  parallelopiped  mutually  bisect 
each  other. 


Prism 

A  G  =  ^-^' 

2 

_b^ 
2* 

h. 

But 

AylBC7=| 

.•.  Prism 

A- 

-G  =  AABC-h, 

i.e.  ] 

pro 

altitude. 

VOLUMES   OF   PYRAMIDS. 


241 


5.  Show  that  the  volume  of  a  triangular  prism  may  be  rep- 
resented by  lialf  the  product  of  one  of  its  lateral  faces  by  the 
perpendicular  distance  of  the  opposite  edge  from  that  face. 

148.  Theorem.  Triangular  pyramids  Jiaving  equivalent 
oases  and  the  same  altitude  are  equivalent* 

Let  A-BDC  and  N-OPQ  represent  the  two  pyramids. 

If  the  altitudes  be  separated  into  any  number  of  equal 
parts,  and  plane  sections  be  passed  through  the  points  of 
division  parallel  to  the  bases,  and  auxiliary  lines  be  drawn 

N 


Fig.  335. 

as  indicated  in  the  figures,  a  set  of  prisms  will  be  formed, 
the  number  of  which  will  be  one  less  than  the  number  of 
parts  into  which  the  altitude  has  been  separated. 

The  plane  sections  at  the  same  distances  from  the 
vertices  will  be  equivalent. 

.-.  EFG=RST, 
IKL  =  VWU,  etc. 

The  three  prisms  in  one  figure  Avill  each  be  equal  to  one 
of  the  three  prisms  in  the  other  figure ;  because  prisms 
having  equivalent  bases  and  the  same  altitude  are  equiva- 
lent (§147). 

*This  Tlieorem  is  what  is  called  a  Lemma.  It  has  been  intro- 
duced to  meet  the  requirements  of  the  succeeding  article. 


242  ELEMENTS  OF  GEOMETRY. 

Let  Yi  and  Z,  represent  the  sum  of  the  three  prisms 
inscribed  in  A-BCD  and  N-OPQ  respectively. 

Because  each  of  the  three  prisms  in  one  set  is  equiva- 
lent to  one  of  the  prisms  in  the  other  set, 

If  each  of  the  previous  divisions  of  the  altitude  be 
bisected  and  new  prisms  be  constructed,  with  their  edges 
parallel  to  AB  and  NO  respectively,  the  two  sets  of 
prisms  (  Fg  and  Zg)  will  be  equivalent  prism  for  prism. 

Z, 

With  this  and  each  further  increase  in  the  number  of 
prisms  (after  the  same  manner)  the  volume  occupied  by 

Y 

the  prisms  will  be  increased;  but  at  each  step  ^=1. 

As  the  number  of  prisms  increases  in  each  pyramid, 
their  sum  approaches  the  volume  of  the  pyramid  as  a 

limit,  and  —  =  1,  where  Y  and  Z  represent  the  volumes 

of  the  pyramids  A-BCD  and  N-OPQ  respectively. 
Hence  Y  =  Z.  Q.  e.  d. 

149.  Problem.  Find  an  eoepression  for  the  volume  of  a 
triangular  pyramid.  e 

Let  A-BCD  represent  the  /  \\ 

figure,  the  volume  of  which 
is  to  be  determined. 


Analysis.  —  If  a  triangu- 
lar prism  as  BCD-AEO  be 
cut  by  the  plane  ACD,  a 
triangular  pyramid  A-BCD  ^^°-  ^^• 

and  a  quadrangular  pyramid  A-ECDO  will  result.     If 


VOLUMES   OF   WEDGES.  243 

the  latter  be  cut  by  tbe  plane  AED,  the  triangular  prism 
will  be  separated  into  three  triangular  pyramids. 

The  pyramids  A-EDG  and  A-EDG  have  equivalent 
bases  and  the  same  altitude,  and  the  pyramids  D-AEG 
and  D-ABC  have  equivalent  bases  and  the  same  altitude. 

If  the  pyramids  having  equivalent  bases  and  the  same 
altitude  are  equivalent  in  volume,  each  will  be  one-third 
of  the  volume  of  the  prism. 

Conclusion.  —By  §  148  D^ABC  =  D-AEC  =  A-DEG. 
.-.  A-BGD  =  i  AEG-BGD, 
or  A-BGD  =  ^h-  BGD  =  ^h-b.        q.  e.  f. 

Exercises.  —  1.  Show  that  the  volume  of  any  pyramid  may  be 
represented  by  — 

o 

2.  Show  that  the  volume  of  a  triangular  pyramid  and  the  volume 
of  any  pyramid  may  be  represented  by  ^  ft  (6  +  47w),  in  which  6  is 
the  base  and  m  is  a  section  paralUl  to  the  base  and  midway  be- 
tween base  and  vertex. 

BH 

V        bh^      bh      \b)[h)' 
.3 

But 

17-  TT3         A3 

etc. 
150.   Definitions.     Figures  (1),  (2),  (3),  and  (4)  represent 


B_ 

b 

-A^- 
a^ 

.     V_ 

V 

a3 

Definitions. 

Fif 

mre 

€B     <0^  S 


(1)  (3)  (3) 

Fig.  837. 

wedges.     The  back  may  be  the  same  length  as  the  edge, 
longer  than  the  edge,  shorter  than  the  edge,  or  it  may  be 


Fig.  338. 


244         ELEMENTS  OF  GEOMETRY. 

only  a  line.  In  the  last  case  the  wedge  is  a  tetraedron, 
two  opposite  edges  of  which  are  considered  back  and 
edge,  and  the  altitude  of  which  is  the  perpendicular  dis- 
tance between  back  and  edge. 

Any  of  the  wedges  represented  in  (1),  (2),  and  (3)  may 
be  converted  into  a  wedge  having  the  form  of  (4),  and  a 
quadrangular  pyramid  ^  c  B 

having  its  vertex  at  one  7iJ/\___     ^-""J^^ 

extremity  of  the  edge.    \- — 7;tA  1l\'iTl  L\  y^      \ 

Problem.     Find  an         \      \    \       /,       ^^     K 

expression  for  the  vol-  \.       h/_ ^^bN. 

ume  of  a  wedge  of  form 

(4). 

Let  ABCD  represent  the  wedge,  h  being  the  perpen- 
dicular distance  between  the  opposite  edges  AB  and  CD. 

Through  C  and  D,  draw  CE  and  DF  parallel  and  equal 
to  AB.     Draw  BE,  BF,  and  EF. 

The  figure  ACDFEB  is  a  triangular  prism  the  volume 
of  which  by  Ex.  5,  §  147,  is  ^  hh. 

A  plane  parallel  to  AB  and  CD,  bisecting  h,  will  inter- 
sect the  wedge  in  a  parallelogram  (m),  whose  sides  are 
one-half  those  of  6. 

Hence  m  =  -,  or  6  =  4m. 
4 

If  W  represent  the  volume  of  the  wedge, 

W  =  ACDFEB  -  (B-CDFE), 

TTT-     hb     hh     hb      1  X,, 

or  TF  =  ^  A  •  4  m.*  q.  e.  f. 

•This  form  might  be  reduced  to  f  hm,  but  this  is  not  so  con- 
venient a  form  as  the  above,  for  purposes  tliat  will  appear  in  the 
next  article. 


PKISMOIDAL   FORMULA. 


245 


Exercise.  —  Show  that  the  volume  of  each  of  the  forms  of 
wedges  (1),  (2),  and  (3),  may  be  represented  by : 

iA(6  +  4n»), 
in  which  b  represents  the  area  of  the  back,  and  m  the  area  of  the 
middle  section. 

151.  Problem.  Deduce  a  formula  for  the  volume  of  a 
figure,  the  upper  and  lower  bases  of  which  are  plane  poly- 
gons in  parallel  planes  and  the  lateral  faces  of  which  are 
triangles  or  quadrangles. 

Let  the  accompanying  figure  represent  a  polyedron,  the 
bases  of  which  are  any  polygons  whatever,  in  parallel 
planes,  and  the  lateral 
faces  triangles. 

Some  of  the  faces  may 
be  quadrangular  plane 
figures,  but  these  may 
always  be  separated  into 
triangles,  having  their 
bases  in  the  perimeter 
of  either  B  or  B'. 

Such  a  figure  may,  by 
passing  planes  through  selected  lines  and  points,  be  cut 
into  pyramids  and  wedges,  the  sum  of  the  bases  of  which 
will  be  B  and  B',  the  sum  of  the  middle  sections,  M,  and 
the  altitude  of  each  sub-figure  be  h,  the  altitude  of  the 
entire  figure. 

The  sum  (P)  of  all  the  pyramids  having  their  bases  in 
B  and  their  vertices  in  B',  will  give, 

P  =  i  ^  (5  -f  4  m),  (Ex.  2,  §  149) 

m  representing  the  part  of  M  which  falls  within  these 
pyramids. 


Fig.  8.39. 


246         ELEMENTS  OF  GEOMETRY. 

The  sum  (p)  of  all  the  pyramids  having  their  bases  in 
B'  and  their  vertices  in  B,  will  give, 

i>  =  |7i(£'  +  4m'), 
m'  representing  the  part  of  M  which  falls  within  this 
second  set  of  pyramids. 

The  remainder  will  be  wedges  of  the  form  (4),  §  150. 

The  sum  (  W)  of  these  wedges,  having  lines  in  B'  and 
B  for  their  upper  and  lower  bases,  will  give, 

The  total  volume  will  be : 

P  =  ^h{B-\-4.m) 
p  =  |7i(B'  +  4m') 

V=^h{B  +  B'-\-AM)  Q.  E.  F. 

Note.  — This  formula,  known  as  the  prismoidal  formula,  is  the 
greatest  of  all  formula  for  the  determination  of  volume. 

Exercises.  —  1.  A  foot  of  lumber  is  a  piece  a  foot  square  and 
an  inch  thick.  Find  the  number  of  feet  of  lumber  in  a  telegraph 
pole  12  in.  square  at  one  end,  4  in.  square  at  the  other,  and  20  ft. 
long. 

2.  Depending  on  §  136,  Ex.  2,  show  that  the  prismoidal  formula 
may  be  used  to  determine  the  volume  of  a  frustum  of  a  cone. 

3.  A  tank,  which  Ls  a  frustum  of  a  coue  in  shape,  is  10  ft.  deep, 
has  a  base  diameter  of  10  ft. ,  and  the  diameter  of  its  upper  base  is 
8  ft.  Find  the  number  of  gallons  it  will  contain,  each  gallon 
being  231  cu.  in. 

152.  Problem.  Show  that  a  truncated  triangular 
prism  equals  the  sum  of  three  pyramids,  the  altitudes  of 
which  will  be  the  altitudes  of  the  three  vertices,  and  the 
bases  of  which  will  be  the  base  of  the  truncated  prism. 


TRUNCATED   TRIANGULAR   PRISM.  247 

The  plane  of  A,  E,  and  C  determines  a  pyramid  having 
Da 


Fig.  S40. 

h'  for  its   altitude   and  ABC  for    its    base. 
E-ABC  =  ^'ABa 

The  plane  of  E,  D,  and  C  determines  the  pyramids 
E-ADC,  and  E-DCF. 

C-ADE  ^  APE  ^AD^h" 
C-ABE     ABE     BE     h'' 

.-.  C-ADE  = -(C-ABE)  =  ^  •  ^ .  ABC  =  —  -  ABC. 
Iv  h'    3  3 

D-ECF  ^  ECF  ^CF  ^  7i"' 
A-EBC     EBC     BE      h'' 

J)'"  h'"    h'  h'" 

.' .  D-ECF  =  ~  (A-EBC)  =  ^.-.  ABC  =  —  •  ABC. 

Q.  E.  D. 

If  V  represent  the  volume  of  the  truncated  triangular 
prism, 

V  =  \-ABC  +  —  -ABC  +  '^-ABG. 

o  o  o 


V  =  ABC 


/h'  +  h"  +  h"\ 


248 


ELEMENTS   OF   GEOMETRY. 


Fi6.  841. 


Problem.  Show  that  the  volume  of  a  truncated  quad- 
rangular prism  equals  the  product  of  its  base  by  the 
average  altitude  of  its  vertices. 

Note.  —  This  fact  is  useful  in  computing  earthwork. 

Let  A,  B,  C,  and  D  represent  points  so  located  that 
their  projection  on  a  horizontal  plane 
will  form  a  square  each  side  of  which 
is,  say,  10  ft.  The  elevation  of  the 
points  A,  B,  C,  and  D  is  determined 
by  taking  their  levels  above  some  as- 
sumed plane,  as  MN. 

The  formula  determines  the  vol- 
ume of  the  figure  ABCDEFHK. 

After  the  excavation,  a  resurvey  and  computation  will 
determine  the  amount  of  dirt  that  has  been  removed. 

153.  Problem.  Show  that  the  volume  of  a  frustum 
of  a  triangular  pyramid  is  equivalent  to  three  pyramids 
each  having  the  altitude  of 
the  frustum  for  its  altitude, 
and  having  for  their  several 
bases  the  upper  base,  the 
lower  base,  and  a  mean  pro- 
portional between  the  two. 

J,  D,  and  C  determine  a 
plane  which  gives  one  of  the 
pyramids,  viz.  D-JAC. 

D,  E,  and  C  determine  a  plane  which  gives  another  of 
the  pyramids,  viz.  C-DEF. 

The  remaining  pyramid,  D-EJC,  is  equivalent  to 
K-EJC,  which  may  be  read,  E-JKC.  {DK  was  drawn 
parallel  to  EJ  and  KE  and  KC  drawn.) 


Fio.  842. 


VOLUME  OF   A   SPHERE. 


249 


If  through  the  point  K,  KH  be  drawn  parallel  to  ACy 
AJKH=AEDF, 
and  (by  §  80,  Ex.  2)  AJKC  is  a  mean  proportional  be- 
tween A  JKH  (=  EDF)  and  A  JAG.  q.  e.  d. 

Expressed  as  a  formula, 


or 


F  =  |(5  +  6  + 


3 


Exercises.  —  1.   Show  that  for  the  frustum  of  any  pyramid, 


^Bb). 


2.    Show  by  the  method  of  limits  that  the  same  formula  may  be 
applied  to  the  determination  of  the  volume  of  a  cone  frustum. 

154.  Pkoblem.     Deduce  a  formula  for  the  volume  of  a 
sphere. 

If  a  cube  be  circumscribed  about  a  sphere,  and  lines  be 
drawn  from  the  vertices  of  the  cube  to 
the  centre  of  the  sphere,  six  pyramids 
will  be  formed,  each  with  the  same  alti- 
tude (R),  and  the  sum  of  the  bases  will 
be  the  surface  of  the  cube.  Represent- 
ing the  sum  of  the  volumes  of  the  pyra- 
mids by  Pi,  and  the  sum  of  the  bases 
by  Si,  we  shall  have, 

3  '       ^1      3 
If  any  number  of  planes  be  passed  tangent  to  the 
sphere,  they  will  cut  off  some  of  the  volume  of  the  cube 
exterior  to  the  sphere. 


Fw.  348. 


250         ELEMENTS  OF  GEOMETRY. 

If,  then,  lines  be  drawn  from  each  of  the  vertices  of  the 
new  circumscribing  polyedron  to  the  centre  of  the  sphere, 
there  will  be  formed  as  many  pyramids  as  the  polyedron 
has  faces. 

Kepresenting  the  sum  of  the  new  pyramids  by  P^,  and 
the  surface  of  the  new  polyedron  by  JS^,  we  will  have, 

'        3   '      S2      3 
If,  again,  any  number  of  planes  be  passed  tangent  to 
the  sphere,  we  shall  have  at  this  stage  of  the  process, 

S,     3 
With  the  next  removal  of  exterior  volume,  the  relations 
between  the  parts  remaining  will  be, 

JSi     3' 

The  same  may  be  continued  indefinitely,  and  each  time 
we  stop  to  consider  the  relations  we  shall  have  an  equa- 
tion of  the  form : 

S„      3 
The  numerator  of  the  first  number  continually  dimin- 
ishes and  approaches  the  volume  of  the  sphere  as  its 
limit ;  and  the  denominator  approaches  the  surface  of  the 
sphere  as  its  limit. 

Hence  if  V  represent  the  volume  of  the  sphere  and  S 
its  surface,  we  shall  have  for  the  relation  between  the 
limits  (see  §  96), 

V^R,      V    ^R 
S      3  '  47ri22      3' 

or  V=UR', 


VOLUMES   OF   SIMILAR   FIGURES. 


251 


Exercises.  —  1.    Apply  the  prismoidal  formula  to  a  sphere. 

2.  Apply  the  prismoidal  formula  to  a  hemi- 
sphere, 

3.  Show  that  the  volumes  of  any  two  spheres 
are  to  each  other  as  the  cubes  of  their  radii,  or  as 
the  cubes  of  any  corresponding  lines. 

4.  The  circumference  of  a  great  circle  of  a 
sphere  being  6  ft.,  find  the  volume. 

5.  Find  the  relative  volumes  of  a  sphere  and 
its  circumscribed  cylinder.  Fig.  S44. 


155.  Theorem.  Tlie  volumes  of  similar  figures  are  to 
each  other  as  the  cubes  on  any  corresponding  lines. 

Let  P  and  p  represent  two 
polyedra  that  are  similar,  i.e. 
the  figures  are  bounded  by  simi- 
lar polygons  which  make  diedrals 
with  each  other  that  in  one  are 
equal  to  those  in  the  other. 

Analysis.  —  These  similar  polyedra  may  be  separated 
into  tetraedra  that  shall  be  similar,  and  if  a  relation 
between  the  volumes  of  similar  tetraedra  be  known,  the 
relation  between  the  volumes  of  similar  polyedra  may 
be  determined. 

Demonstration.  —  (By  §  149,  Ex.  3.)  Similar  pyramids 
are  to  each  other  as  the  cubes  on  any  corresponding  lines. 
And  if  the  pyramids  be  represented  by  P,,  pi,  P^,  p^, 
etc.,  and  the  corresponding  lines  by  Xj,  ?i,  L^,  I2,  etc., 
we  will  have, 

Pi     L\   P2     L' 


Pi        ^i     i>2        l^: 


-,  etc. 


252  ELEMENTS   OF  GEOMETRY. 

But  by  reason  of  the  fact  that  in  similar  figures  cor- 
responding lines  are  proportional, 

—  =  —  =  —?  etc.   =— . 

il  I2  1$  I 

Eecalling  the  fact  that  if  proportions  have  common 
ratios  they  may  be  added  term  by  term  (see  §  79), 

A  +  Pg  +  P3  +  etc.  ^  P_^X» 
i>i+i)2  +  i?3  +  etc.       p      P' 

which  establishes  the  theorem  as  far  as  similar  polyedra 
are  concerned. 

There  are,  however,  volumes  bounded  by  curved  sur- 
faces alone,  and  volumes  bounded  by  surfaces  some  of 
which  are  curved  and  some  of  which  are  plane. 

Such  surfaces,  provided  they  be  similar,  as  stated  in 
the  announcement  of  the  theorem,  may  have  inscribed 
within  them,  or  circumscribed  about  them,  similar  poly- 
edra, which  may  be  so  constructed  that  they  shall  have 
at  least  one  pair  of  corresponding  lines  which  in  the  sub- 
sequent increase  in  the  number  of  faces  shall  remain 
unchanged.  These  similar  polyedra  are  to  each  other  as 
the  cubes  on  any  similar  lines. 

If  Pj  and  pi  represent,  say,  the  circumscribing  poly- 
edra, and  L  and  I  a  pair  of  corresponding  lines  that  shall 
remain  unchanged,  we  shall  have. 

If  the  number  of  faces  be  increased  by  planes  tangent 
to  the  volumes,  we  shall  have, 

P2^X» 


VOLUIVIES   OF   SIMILAR   FIGURES.  253 

At  each  step  likewise, 

and  at  each  step  we  shall  have  P„  and  j>„  representing 
volumes  which  are  diminishing  and  which  are  approach- 
ing fixed  voliunes  as  their  limits. 

By  §  96  we  arrive  at  the  conclusion  that 
V    U 

Note, — The  principle  just  deduced  is  the  third  of  tlie  tliree 
great  principles  of  elementary  geometry  heretofore  alluded  to. 

Exercise.  —  Show  how  a  numerical  representative  of  a  rectan- 
gular parallelepiped  may  be  determined. 

Suggestion.  — Pursue  the  method  of  §  57. 


CHAPTER  XIV. 

An  Introduction  to  the  Study  of  the  Plane 
Sections  of  a  Right  Circular  Cone. 


166.  If  we  conceive  of  a  plane  tangent  to  a  right 
circular  cone  (§  136),  and  the  plane  through  the  element 
of  tangency  and  the  axis  be  represented  by  the  page  on 
which  we  have  the  figure, 
these  planes  will  be  per- 
pendicular to  each  other 
(§§  107,  112). 

If  a  plane  be  passed  paral- 
lel to  the  tangent  plane,  it 
will  intersect  one  nappe  of 
the  cone  and  not  the  other. 
This  line  of  intersection  will 
be  a  plane  curve  which  is 
called  a  parabola. 

The  curve  will  change  its 
form  depending  upon  the 
distance  of  the  secant  plane  from  the  tangent  plane. 

There  are  many  properties  peculiar  to  the  parabola, 
a  few  of  which  will  be  deduced  in  this  chapter. 

Theorem.  The  ratio  of  the  distances  of  every  point  of 
a  parabola  from  a  certain  point  in  its  plane  and  from  a 
certain  straight  line,  also  in  the  jilane,  equals  1. 

254 


Fio.  340. 


THE  PARABOLA.  255 

Within  the  cone,  tangent  to  the  cone  and  also  tangent 
to  the  secant  plane,  conceive  a  sphere  to  be  constructed. 
It  will  be  tangent  to  the  cone  on  one  of  its  small  circles, 
the  plane  of  which  will  be  perpendicular  to  the  axis  of 
the  cone  ;  and  for  that  reason,  perpendicular  to  the  plane 
on  which  the  figure  is  represented. 

The  plane  of  the  small  circle,  and  the  plane  which 
intersects  the  surface  of  the  cone  in  a  parabola,  are  both 
perpendicular  to  the  plane  of  the  picture ;  and  so  (§  111, 
Ex,  4)  their  line  of  intersection  {DM)  is  perpendicular 
to  the  plane  of  the  picture. 

Let  P  be  any  point  on  the  parabola.  Join  it  to  F  (the 
point  of  tangency  of  the  sphere  and  the  secant  plane), 
to  V  (the  vertex  of  the  cone),  and  draw  PM  A.  to  DM, 

Through  P  pass  a  plane  perpendicular  to  the  axis  of 
the  cone ;  it  will  intersect  the  surface  of  the  cone  in  the 
O  TPTi,  and  the  plane  of  the  picture  in  YT^. 

The  plane  of  the  parabola  intersects  the  plane  of  the 

picture  in  QD.  

PF=PR 

(tangents  to  a  sphere  from  P)  ; 

PR  =  TN 

(portions  of  elements  between  parallel  planes)  ; 

TN=TA+AN, 

TA  +  AN^  QA  +  AD 

(AAQT  and  AND  are  isosceles) ; 

QA  +  AD  =  PM 

(opposite  sides  of  a  rectangle  are  equal). 

.-.  PF=P3I, 

PF 
—  =  1.  Q.E.D. 


256 


ELEMENTS   OF  GEOMETRY. 


Note. — Sections  of  a  cone  made  by  planes  not  parallel  to  a 
tangent  plane  will  be  considered  after  a  few  of  the  simpler  proper- 
ties of  the  parabola  have  been  deduced. 

Most  of  the  properties  of  the  Conic  Sections  are  best  developed 
by  the  methods  of  Analytic  Geometry  and  Calculus  ;  but  because 
of  their  importance  in  Nature  and  in  Art,  a  few  of  the  simpler 
properties  are  here  deduced. 

157.  Definitions.  By  reason  of  the  property  deduced  in 
the  preceding  article,  the  parabola  is  often  described  as : 
The  locus  of  a  point  moving  in  a  plane  so  that  the  ratio 
of  its  distances  from  a  fixed  point  and  a  fixed  straight  line 
equals  1. 

The  fixed  point  is  called  the  focus. 

The  fixed  straight  line  is  called  the  directrix. 

The  ±  FD  is  called  the  axis,  because  the  curve  is 
symmetrical    with    respect    to 
this  line :  P^Q  =  PQ,  as  may  be 
show  from  the  figure  in  §  156. 

A  is  called  the  vertex.  QP  is 
called  an  ordinate. 

The  double  ordinate  through 
F  is  called  the  parameter  or  latus 
rectnm. 

Exercise.  —  Show  that  the  ordinate  at  the  focus  equals  FD. 

158.  Problem.  Having  given  the  fixed  point  and  tlie 
fixed  line  in  a  plane,  to  construct  the  parabola. 

Let  AB  be  the  given  line 
and  F  the  given  point. 

The  analysis  suggests  the 

Construction. — Draw  a  num- 
ber of  lines  parallel  to  AB  on 
the  side  that  F  is. 

With  jP  as  a  centre  and  the  fio.  848. 


IW 

^ — • 

D 

AIF 

f 

P 

Q 

My 

\ 

^ 

Pi 

Fig.  847. 


THE   PARABOLA. 


257 


distance  of  the  first  line  from  AB  as  a  radius,  construct 
an  arc,  intersecting  the  first  line. 

These  two  points  (C  and  D)  of  intersection  are  two 
points  on  the  parabola. 

Repeat  the  process,  using  the  second  line  in  the  same 
way  as  the  first  was  used,  thus  determining  the  points 
E  and  G. 

In  the  same  way  H  and  K  and  any  number  of  points 
may  be  determined. 

Through  these  points  draw  a 
smooth  curve;  it  will  approxi- 
mate closely  to  the  required 
parabola. 

A  mechanical  construction  may 
be  affected  by  using  the  edge  of 
a  ruler  as  the  directrix,  against 
which  one  perpendicular  side  of 
a  right  triangle  is  caused  to  slide,  fig.  349. 

while  against  the  other  perpendicular  side  the  marking 
point  P  presses  a  string,  one  end  of  which  is  secured  at  F 
and  the  other  so  fastened  that 


KP+PF=KM. 

P  will  be  a  point  on  a  parabola,  because  in  any  proper 
position 

PF=PM. 


Exercise. — Construct  parabolas,  liaving  given  the  distance  of 
Ffrom  the  directrix,  1,  3,  10,  and  20  centimetres. 


258 


ELEMENTS   OF   GEOMETKY. 


159.  Theorem.  If  a  line  be  drawn  to  the  focus  from 
the  intersection  of  a  secant  with  the  directrix,  it  will  bi- 
sect the  exterior  angle  of  the  triangle  formed  by  the  chord 
and  the  focal  radii. 

P'W,  and  P"W  being  parallel, 
P'M"      BP' 


P"M" 

~  RP' 

But 

P'M' 

=  FP', 

and 

P"M" 

^FP. 

.   FP' 

BP> 

•  •  ppT, 

~EP" 

Fig.  350. 


Therefore  (§69,  Prob.  7)  the  line  FR  bisects  the 
exterior  angle  of  the  A  P'FP". 

Corollary.  If  P'  and  P"  should  coincide  at  P,  PR 
would  be  a  tangent  and  FR  would  be  perpendicular  to  the 
focal  chord  through  F  and  P. 

160.  Theorem.  A  tangent  to  a  parabola  bisects  the 
angle  formed  by  PF  and  PM. 

The  analysis  suggests  that  we  establish  the  equality 
of  the  AMPR  and  FPR 
if  possible. 

Demonstration.  —  These 
triangles  are  equal  because 
each  has  a  right  angle;  a 
side  adjacent  to  the  right 
angle  in  each  equal ;  and 
the  hypothenuse  common. 

.  • .  Z  MPR  =  Z  FPR.  Fio.  361. 


THE   PARABOLA. 


269 


Note. — One  of  the  important  uses  to  which  this  property  is 
put  is  the  construction  of  parabolic  reflectors. 
/.8FJ=^FPr. 

And  since  the  angle  of  incidence  equals  the  angle  of  reflection, 
any  ray  of  light  emanating  from  F  will  be  reflected  to  a  line  paral- 
lel to  the  axis. 

The  parabolic  reflector  is  made  by  revolving  a  parabola  on  its 
axis.  As  a  geometric  figure  it  is  called  a  paraboloid  of  revolution. 
If  rays  of  light  should  enter  the  parabolic  reflector  parallel  to  the 
axis,  they  would  all  be  reflected  to  the  focus. 

Exercises.  —  1.  Show  that  the  distance  from  the  focus  to  the 
point  of  tangency  equals  the  distance  from  the  focus  to  the  foot  of 
the  tangent. 

2.  Show  that  if  FM  be  drawn,  it  will  be  perpendicular  to  the 
tangent,  and  will  be  bisected  by  the  tangent. 

3.  Show  that  the  locus  of  the  foot  of  a  perpendicular  from  the 
focus  to  the  tangent  will  be  the  tangent  at  the  vertex  of  the  curve. 

4.  Show  that  the  projection  on  the  axis  of  the  segment  of  the 
tangent  is  bisected  at  the  vertex.  (This  projection  is  called  the 
sub-tangent.) 

161.   Definitions.     A  normal  is  a  Hue  perpendicular  to  a 
tangeut  aud  passiug 
through  the  point  of 
tangency. 

A  sub-normal  is  the 
projection  on  the  axis 
of  the  segment  of 
the  normal  included 
between  the  point  of  ^'«-  352. 

tangency  and  the  axis.     Thus  QN  is  the  sub-normal. 

Theorem.     The  sub-normal  of  a  xmrahola  is  constant, 
and  equals  the  distance  of  the  focus  from  the  directrix. 
APQN=AMDF. 
.-.  QN=DF. 


260 


ELE]MENTS   OF   GEOMETRY. 


Exercise.  —  1.  Show  how,  in  three  ways,  to  draw  a  tangent  to 
a  parabola  at  a  given  point  of  the  curve. 

162.  Theorem.  If  from  the  point  of  intersection  of 
two  tangents  a  straight  line  he  drawn  parallel  to  the  axis  of 
a  parabola,  it  will  bisect  the  chord  joining  the  points 
of  tangency. 

Analysis.  —  If  P^B  =  BPo, 
MiMs  will  equal  3/33/3  and 
the  AM1TM2  will  be  isosce- 
les. 

Demonstration. — Since  IPi 
is  the  perpendicular  bisector  fig.  353. 

of  FMi  (Ex.  2,  §160),  IF  =  nil. 

Also  since  IP^  is  the  perpendicular  bisector  of  FM2, 
IF  =  IMo. 
.-.  IM^  =  IM^ 
and  AM1IM2  is  isosceles. 

.-.  3/11/3  =  3/33/2 
and  PiB  =  BP^  q.  e.  d. 

Corollary.  If  at  the  point  of  intersection  of  IB  icith  the 
curve  a  tangent  be  draicn,  it  will  be  parallel  to  the  chord 
PiP-i,  and  the  segment  IB,  which  bisects  the  chord  PiP^,  is 
itself  bisected  by  the  curve. 

Analysis.  —  If  a  tan- 
gent at  the  point  C  is 
parallel  to  P1P2,  and  if 
it  bisects  IB,  it  will  also 
bisect  /Pi  and  IP^. 

Demonstration.  —  Let 
IJo  represent  the  tangent  at  C,  intersecting  7Pi  and  IP.,. 


Fig.  351. 


THE   PARABOLA.  261 

By  the  theorem,  a  straight  line  through  Jj,  parallel  to 
the  axis,  bisects  CPi. 

IiIIi  is  parallel  to  IC  (each  being  parallel  to  the  axis). . 

.•.  Jj  is  the  middle  point  of  IPi. 

In  the  same  manner  it  is  shown  that  I^  is  the  middle 
point  of  IPo. 

I1I2,  bisecting  two  sides  of  the  A IP1P2,  is  parallel  to  the 
side  P1P9  and  bisects  IB.  Q.  e.  d. 

Problem.  Show  how  to  construct  an  arc  of  a  parabola 
having  given  a  pair  of  tangents  and  the  points  of  tangency. 

C  --^^  ~"^--,:..^ 


Fig.  365. 

The  analysis  suggests  the  following : 

Join  P1P2 ;  bisect  it  at  B. 

Join  BI;  bisect  it  at  A. 

Join  APi ;  bisect  it  at  C. 

Draw  CE  parallel  to  BI;  bisect  it  at  7). 

Join  AP., ;  bisect  it  at  //. 

Draw  //,/  parallel  to  BI;  bisect  it  at  K. 

The  points  Pj,  I  J,  A,  K,  and  P^,  are  five  points  on  the 
required  curve. 

As  many  points  as  are  desired  may  be  found  in  the 
same  way,  and  the  curve  traced  throu.gh  them  will  be 
the  arc  of  the  parabola  required. 

Note. — The  method  described  in  the  above  problem  is  one  of 
the  processes  frequently  used  for  constructing  parabolic  railroad 
curves. 


262 


ELEMENTS   OF   GEOMETRY. 


163.  Problem.  Show  that  the  area  included  between 
an  arc  and  a  chord  of  a  parabola  is  two-thirds  the  area 
of  the  triangle  formed  by  the  chord  and  the  tangents  at 
its  extremities. 

Let  P1P2  represent  the  chord,  P1CP2  the  arc,  I  Pi  one 
tangent,  and  IP^  the  other. 

B  is  the  middle  point  of  P1P.2.  Join  IB.  Draw  a 
tangent  at  C  and  draw  the  chords  GP^  and  CP.,. 

The  area  of  the  AP1CP2  is  double  the  area  of  the 
A I1II2 ;  the  base  P1P2  is 
twice  the  base  I^L  and 
the  altitudes  are  the 
same.  The  AP1CP2  is 
said  to  be  inscribed  and 
the  A  I1II2  is  said  to  be 
escribed. 

If  at  K  and  at  H  tan- 
gents be  drawn  to  the 
curve,  and  chords  be 
drawn  from  Kto  Pi  and 
to  C,  and  also  from  H  to 
C  and  P2,  we  shall  have 

an  inscribed  convex  polygon,  and  an  escribed  re-entrant 
polygon.  The  additions  to  the  area  of  the  inscribed 
triangle  (which  have  formed  the  inscribed  j^olygon  of 
5  sides)  are  double  the  additions  to  the  area  of  the 
escribed  triangle  (which  have  formed  the  escribed  poly- 
gon of  5  sides). 

For  these  reasons  the  inscribed  polygon  of  5  sides  has 
twice  the  area  of  the  escribed  polygon. 

If  at  the  points  of  intersection  of  the  new  tangents 
with  those  previously  constructed,  lines  be  drawn  parallel 


Fio.  856. 


THE   PARABOLA.  263 

to  the  axis,  tliey  will  determine  points  on  the  curve, 
at  which  if  tangents  be  drawn,  and  to  which  if  cords  be 
drawn  from  adjacent  vertices  previously  determined, 
inscribed  and  escribed  polygons  of  9  sides  would  be 
formed,  each  having  larger  areas  than  the  polygons  of 
5  sides.  The  increase  in  the  inscribed  polygon  is  double 
that  in  the  escribed  polygon. 

The  total  area,  then,  of  the  inscribed  polygon  will  be 
twice  that  of  the  escribed  polygon. 

Let  Ai,  Ao,  Ag,  A^,  •••  represent  the  areas  of  the  in- 
scribed polygons  of  3, 5,  9, 17,  etc.,  sides ;  let  a^,  a^,  a^,  di,--' 
represent  the  areas  of  the  corresponding  escribed  poly- 
gons ;  and  let  A  and  a  represent  the  limits  toward  which 
we  approach  as  the  number  of  sides  is  increased.  The 
nature  of  the  process  and  the  steps  already  taken,  show 
that 

Ai 9     A2 9     Ag 9  A„ 9 

cii  ttg  ttg  a„ 

The  ratio  as  n  increases  is  always  2 ;  hence  it  will  be 

2  at  the  limit,  i.e.  —  =  2,  or 
a 

A 

"=2- 

At  each  step  An  approaches  as  its  limit  the  area 
bounded  by  the  first  chord  and  its  subtended  arc  ;  and  a„ 
approaches  as  its  limit  the  area  bounded  by  the  initial 
tangents  and  the  given  arc. 

If  T  represent  the  area  of  the  A  P1/P9, 

A-{-a=T,  or  A-\-~=T; 

-T-  ,  ^2 

f  A  =  T,  or  A  =  ^  T.  q.  e.  f. 


264         ELEMENTS  OF  GEOMETRY. 

liemark.  —  The  area  of  the  A  T  equals  a  parallelogram  having 
CB  and  BP<i  as  its  adjacent  sides.  If  the  chord  be  perpendicular 
to  the  axis,  the  parallelogram  would  be  a  rectangle. 

It  is  interesting  to  note  that  we  can  convert  an  area  bounded  by 
a  chord  and  an  arc  of  a  parabola  into  an  equivalent  square  ;  a  thing 
we  are  unable  to  do  in  the  case  of  one  bounded  by  a  chord  and  a 
circular  arc. 

Note. — The  parabola  is  one  of  the  most  interesting  of  curves. 

The  majority  of  comets  move  along  parabohis  with  the  sun  at 
the  focus. 

The  cables  of  a  suspension  bridge,  when  sustaining  their  own 
weight  alone,  form  a  curve  called  a  catenary,  but  when  loaded  by 
a  uniformly  distributed  weight  (as  when  the  road-bed  is  attached), 
the  catenary  is  changed  to  a  parabola. 

A  waterway  in  which  the  water  will  flow  with  unifoi-m  velocity, 
whatever  its  depth  may  be,  will  have  a  parabola  for  its  cross- 
section. 

The  parabolic  arch  is  the  proper  one  to  construct  where  a 
uniformly  distributed  load  is  to  be  sustained. 

Both  ends  of  a  hen's  egg  are  paraboloids  of  revolution. 

The  flight  of  a  projectile  approximates  closely  to  a  parabola. 


PARTICULAR  CASES. 

164.  If  the  plane  which  intersects  the  surface  of  a 
right  circular  cone  in  a  parabola  be  moved  parallel  to 
itself  until  it  coincides  with  the  tangent  plane  to  which 
it  is  parallel,  the  parabola  will  degenerate  to  a  straight 
line,  which  is  said  to  be  a  particular  case  of  a  parabola. 
If  the  plane  be  further  nioved  in  the  same  direction,  the 
section  will  be  a  parabola  in  the  other  nappe. 

If  we  conceive  the  elements  of  a  right  circular  cone  as 
increasing  the  angle  which  they  make  with  a  given  base, 
the  cone  ^vill  approach  a  cylinder.  When  the  angle  that 
the  elements  make  with  the  base  shall  equal  90°,  the  cone 


THE  ELLIPSE. 


266 


will  have  become  a  cylinder  (a  particular  case  of  a  cone), 
and  a  section  made  by  a  plane  parallel  to  a  tangent  plane 
will  be  two  parallel  straight  lines,  another  particular  case 
of  a  parabola. 

If  we  conceive  the  elements  of  a  right  circular  cone  as 
decreasing  the  angle  which  they  make  with  a  given  base, 
the  cone  will  approach  the  plane  of  its  base,  the  parabola 
will  approach  a  straight  line ;  and  when  the  cone  shall 
have  become  a  plane,  the  parabola  will  have  become  a 
straight  line. 

THE   ELLIPSE. 

165.  If  a  plane  be  passed  through  a  right  circular  cone 
intersecting  all  the  elements  in  one  nappe,  the  line  of  in- 
tersection is  an  ellipse. 

The  section  is  a 
closed  curve ;  for  if  a 
point  on  the  section  be 
moved  along  the  curve 
in  the  same  sense,  it 
will  return  to  its  in- 
itial position. 

Two  spheres  may  be 
inscribed  "wathin  the 
cone  that  shall  at  the 
same  time  be  tangent 
to  the  secant  plane  (as 
at  F  and  F '),  and  tan- 
gent to  the  surface  of 
the  cone  (in  circles,  the 
planes  of  which  are  perpendicular  to  the  axis  of  the  cone). 

Let  the  plane  of  the  paper  be  the  plane  through  the 


Fig.  85T. 


266         ELEMENTS  OF  GEOMETRY. 

axis  of  the  cone  and  perpendicular  to  the  secant  plane. 
It  will  intersect  the  planes  of  the  small  circles  of  tan- 
gency  in  lines  DM  and  D'M',  perpendicular  to  the  plane 
of  the  picture.* 

Let  P  represent  any  point  of  the  section,  a,nd  PQT  a 
plane  through  P,  perpendicular  to  the  axis  of  the  cone. 

PF^PR=  TJV  =TA  +  AN, 
PM=QD=QA+AD. 

From  the  similar  A  AQT  and  ADN,  we  have, 

IA=,9A.     TA-\-AN^QA-\-AD. 

AN     AD'         AN  AD      ' 

TA  +  AN^AN 

QA  +  AD     AD 

.  PF  ^AN^AF 

PM     AD     AD 

AN  • 

— —  is  constant  for  this  position  of  the  secant  plane  and 

is  <  1,  because  in  the  A  AND,  AN  is  opposite  a  smaller 
angle  than  AD  is.     Hence  we  have  the 

Theorem.  In  the  plane  of  an  ellipse  there  is  a  point 
(called  a  fociis),  and  a  straight  line  (called  a  directrix j,  the 
distances  from  which  to  any  point  of  the  curve  will  have  a 
fixed  ratio,  and  that  ratio  will  be  less  than  1. 

Exercises.  —  1.   Show  that  ^^-^-  =  :i2-i^. 
PM'     A'D' 

2.   Show  that  ^^  =  — ,  and  hence  ~  =  ^^  (§65,  Ex.  2). 
A'ly     AD  PM      PM'  ^^  ' 

8.    Show  that  Pq  =  QP. 

*It  is  expressly  nnderstnod  that  the  sketches  which  serve  as 
figures  for  purposes  of  determining  relations  between  parts  are 
not  perspective  drawings. 


THE   ELLIPSE. 


267 


166.  Theorem.  Tlie  sum  of  the  distances  of  any  point- 
of  an  ellipse  from  two  certain  fixed  points  in  its  plane  is 
constant. 

By  the  figure  of  the  preceding  article,  it  is  seen  that 
PF  =  PR  and  PP  =  R'P. 
. • .  PF  +  PF'  =  B'E  (a.  constant).  Q. e. d. 

Problems.  —  1.   Show  that  AF  =  A'F. 


Fig.  358. 


Since  PF  +  PF'  is  a  constant,  the  sum  will  not  be  altered  by 
taking  P  in  a  particular  position.  K  it  be  moved  to  A  and  then  to 
A',  we  shall  have, 

AF  +  AF  =  A'P  +  A'F, 

or  AF+AF+  FF'  =  A'P  +  A'F'  +  FF', 

or  2  AF-2  A'F, 

AF  =  A'F. 
2.    Show  that  PF  +  PF'  =  AA'. 

Definilions.  — Tlie  points  F  and  F  are  called  the  foci. 

The  lines  Dilf  and  D'M'  are  called  directrices. 

A  figure  is  symmetrical  with  respect  to  a  line  if  a  perpendicular 
to  the  line  at  any  point  intersects  the  figure  at  equal  distances  on 
either  side  of  the  line.  The  perpendicular  bisector  of  AA'  is  called 
the  minor  axis. 

The  segment  AA'  through  the  foci  is  called  the  major  axis  and 
is  a  line  of  synnnetry. 

An  ellipse  may  be  defined  as  the  loc7is  of  a  point  moving  so  that 
the  ratio  of  its  distances  from  a  fixed  point  and  from  a  fixed 
straight  line  is  constant  and  less  than  1. 


268  ELEMENTS   OF   GEO:viETKY. 

An  ellipse  may  also  be  defined  as  the  locus  of  a  point  moving  so 
that  the  sum  oi  its  distances  from  two  fixed  points  is  constant. 

Exercises. — 1.  Assume  a  point  and  a  straight  line  and  construct 
by  points  an  ellipse  having  the  ratio  -=—  =  — 

2.  Two  points  are  10  centimetres  apart.  Construct  by  con- 
tinuous motion  an  ellipse  the  major  axis  of  which  is  12  centimetres. 

3.  From  any  point  within  an  ellipse  the  sum  of  the  distances  to 
the  foci  is  less  than  the  major  axis,  and  from  any  point  without  an 
ellipse  the  sum  of  the  distances  to  the  foci  is  greater  than  the 
major  axis. 

'  Q' 


Fig.  359. 

Note.  — There  are  so  many  ways  of  constructing  an  ellipse  with 
reasonable  accuracy  that  there  is  no  excuse  for  making  up  a  com- 
bination of  arcs  of  circles  and  calling  the  figure  an  ellipse. 

167.  Theorem.  If  a  straight  line  be  drawn  through  a 
point  of  an  ellipse  so  as  to  make  equal  angles  tvith  the 
focal  radii  to  the  same  point,  it  ivill  be  a  tangent  to  the  curve. 

A  tangent  is  the  limiting  position  toward  wliich  a 
secant  approaclies  as  the  two  points  of  intersection 
approach  coincidence. 
When  the  secant  shall 
have  become  a  tangent, 
all  of  its  points  excejit  the 
point  of  contact  will  lie 
without  the  curve.  kX^. 


THE   ELLIPSE.  269 

Produce  F'P  so  that  PH  shall  equal  PF;  making 
PHF  an  isosceles  triangle.  Through  P  draw  RQl.  to 
FH.,  it  will  make  equal  angles  with  PF  and  PF'. 

If  i?Q  be  a  tangent,  every  point  except  the  point  of 
contact  P  must  be  shown  to  be  exterior  to  the  ellipse. 

Let  R  represent  any  point  other  than  P. 

F'B  +  RH  >  F'H.  (§  167,  Ex.  3) 

But  RH  =  RF  and  F'H  =  AA'. 

.-.  F'R  +  RF>AA'. 

Hence  the  point  R,  which  may  be  any  point  on  the  line 
other  than  P,  is  exterior  to  the  curve,  and  ^Q  is  a  tangent. 

Note.  —  If  an  ellipse  be  revolved  about  its  major  axis,  a  prolate 
ellipsoid  of  revolution  is  generated.  If  revolved  about  its  minor 
axis,  an  oblate  ellipsoid  of  revolution  is  formed.  The  earth  is 
approximately  an  oblate  ellipsoid  of  revolution. 

168.  Theorem  I.  An  ellipse  is  determinable  when  Us 
axes  are  given. 

We  have  seen  that  an  ellipse  is  the  locus  of  a  point 
which  moves  so  that  the  sum  B 

of  its  distances  from  F  and 
F'  equals  AA'  (2  a).  ^.i 

When  the  moving  point  is       \^ 
at  5, 

BF  =  BF'  =  a.  Fig.  361. 

If,  then,  the  segments  which  are  to  be  axes  are  placed 
so  as  to  mutually  bisect  each  other  at  right  angles,  the 
foci  may  be  determined  by  constructing  an  arc  of  a 
circle  with  B  as  centre  and  a  as  radius.  The  foci  and 
the  major  axis  determine  the  ellipse  (§  166) ;  therefore 
the  axes  do. 


270 


ELEMENTS   OF  GEOMETRY. 


Within  the 


Theouem  II.     A  section  oblique  to  the  elements  of  a 
right  circular  cylinder  will  be  an  ellipse. 

Let  APA'  represent  the  oblique  section, 
cylinder  and  tangent  to  the  secant 
plane,  conceive  spheres  to  be  inscribed 
as  indicated  in  the  figure;  they  will 
be  tangent  to  the  secant  plane  at  two 
points,  as  F  and  F'. 

Draw  PF,  PF',  and  RR'. 

PF=  PR, 
PF'  =  PR', 

PF=  PF'  =  RR'  (a  constant). 
.'.  the  section  is  an  ellipse.       q.  e,  d.  F'g-  3C2. 

Corollary.     Any  right  section  is  a  circle;  so  that  an 
ellipse  may  be  projected  into  a  circle. 

169.   Problem.     If  a  and  b  represent  the  semi-axes 
of  an  ellipse,  show  that  the  area  is  represented  by  tt  ab. 

Let  APA'  represent  an  ob- 
lique section  of  a  right  circvdar 
cylinder  and  KHK'  a  right  sec- 
tion of  the  same  passed  through  -^ 
the  middle  point  of  jLA'. 

BB'  will  be  perpendicular  to  a! 
AA'   and  will    be    the    minor 
axis    of    the    ellipse,   as   well 
as  being  the  diameter  of  the 
O  KHK'. 

From  P  draw  PQ 1.  to  CB  and  PH  perpendicular  to 
the  plane  of  the  circle. 

QH  will  also  be  perpendicular  to  CB  (§  112). 


Fio.  363. 


THE   ELLIPSE.  271 

The  A  PQH  and  ACK  are  similar. 

QP  _CA_a 
QH~CK~b' 

If  the  point  P  should  move  from  5  to  ^  to  A',  the 
segment  PQ  remaining  perpendicular  to  BB'  would  gen- 
erate half  the  area  of  the  ellipse ;  and  the  segment  HQ 
would  generate  half  the  area  of  the  circle. 

Each  line  segment  would  remain  perpendicular  to 
BB',  would  move  the  same  distance,  and  the  ratio  of 
their  lengths  would  remain  the  same,  therefore 

Half  the  Ellipse  _  a 

Half  the  Circle  ~  V 

Ellipse      a 

or  i —  =  — 

Circle       b 

Ellipse  _  a 

Ellipse  =  - —  =  TT  ah.  q.  e.  f. 

b 

Exercises.  —  1.  Deduce  the  same  expression  for  the  area  of  an 
ellipse  by  inscribing  a  polygon  within  the  ellipse,  comparing  its 
area  with  the  area  of  the  projected  polygon  in  the  circle,  and  then 
by  the  theory  of  limits  determine  the  relation  between  the  areas 
of  the  ellipse  and  the  circle. 

2.  Show  that  an  ellipse  has  a  centre,  i.e.  a  point  through  which, 
if  chords  be  drawn,  they  will  be  bisected. 

3.  Show  that  if  a  line  be  drawn  from  the  centre  of  the  ellipse 
to  the  point  Q,  in  the  figure  of  §  167,  its  length  will  be  a.  Hence 
the  locus  of  the  foot  of  a  perpendicular  from  a  fociLs  to  a  tangent 
is  the  circumference  of  a  circle  on  the  major  axis  as  a  diameter. 

This  circle  is  called  the  director  circle. 

Note.  —  The  earth's  meridians  are  ellipses,  the  axis  of  the 
earth  being  the  minor  axis  of  each  ellipse. 


272  ELEMENTS   OF   GEOMETRY. 

It  is  this  fact  which  gives  rise  to  the  statement  that  the  Missis- 
sippi River  runs  up  liill.  Its  mouth  is  further  from  the  centre  of 
the  earth  than  its  source. 

The  locus -of  the  earth  in  its  annual  motion  about  the  sun  is 
approximately  an  ellipse  with  the  sun  at  one  focus. 

If  a  light  (as  an  electric  arc-light)  were  placed  at  one  focus  of 
a  prolate  ellipsoid ,  all  rays  reflected  from  the  surface  would  meet 
at  the  other  focus.     (Established  by  §  167. ) 

Whispering  galleries  are  also  made  which  depend  upon  this 
principle. 

PARTICULAR   CASES. 

170.  If  a  plane  which  intersects  the  surface  of  a  cone 
in  an  ellipse  be  moved  parallel  to  itself  until  it  passes 
through  the  vertex,  the  ellipse  will  degenerate  to  a  point. 
If  the  plane  be  moved  further,  an  ellipse  in  the  other 
nappe  will  be  the  result. 

If  a  plane  which  cuts  an  ellipse  from  a  right  circular 
cone  be  rotated  about  some  axis  toward  the  position  of 
being  perpendicular  to  the  axis,  the  foci  will  approach 
each  other,  and  when  the  plane  becomes  perpendicular 
to  the  axis,  the  foci  and  centre  will  coincide  and  the 
ellipse  will  become  a  circle. 

THE   HYPERBOLA. 

171.  Definitions.  If  a  plane  intersect  a  right  circular 
cone  and  make  with  the  plane  of  a  circular  section  an 
angle  greater  than  that  made  by  the  elements  of  the  cone 
with  the  same  section,  it  will  intersect  both  nappes  of 
the  cone. 

The  line  of  intersection  is  called  a  hyperbola.  In  gen- 
eral it  is  in  two  branches,  neither  of  which  close. 


THE   HYPERBOLA. 


273 


The  two  brandies  are,  however,  superimposable,  as 
will  be  shown  later  in  this  chapter. 

P,  P',  P'"  represent  three  points  of  the  secant  plane,  S 
and  jS'  represent  spheres  tangent  to  the  cone  and  to  the 
secant  plane. 


Fig.  864. 


Let  P  be  any  point  in  one  branch.  Throngh  P  pass  a 
plane  perpendicular  to  the  axis  of  the  cone,  giving  the 
circle  PTP". 

PV  is  an  element  of  the  cone,  tangent  to  the  spheres 
at  B  and  R'. 

PF  and  PF'  are  straight  lines  to  the  points  of  tangency 
of  the  inscribed  tangent  spheres. 


274  ELEMENTS    OF   GEOMETRY. 

DM  and  D'M'  are  the  intersections  of  the  secant  plane 
with  the  planes  of  the  circles  of  taugency  of  the  spheres 
and  the  cone. 

QQ'  is  the  intersection  of  the  secant  plane  with  the 
plane  of  the  picture. 

PF=  PR  =  TN=  TA  +  AN, 
PM=  QD  =  QA  +  AD. 

.    PF  ^  TA  +  AN 

'  '  PM     QA  +  AD 

But  A.AQT  and  AND  are  similar. 

.    TA      QA        TA  +  AN     QA  +  AD 

.. = -^ — ,  or ' =  -^ J 

AN     AD'  AN  AD 

or  TA±AN^AN 

QA  -f-  AD     AD 

.    PF^AN 
PM     AD 

For  this  particular  section,  AN  and  AD  are  fixed ;  ar^ 
AN,  being  opposite  a  greater  angle  than  AD  in  the 
A  AND,  is  greater  than  AD. 

T*F 

Hence is  constant  and>l. 

PM 

Note.  — Tliis  ratio,  which  for  the  Parabola  =  1,  for  the  Ellipse 
<1,  and  for  the  Hyperbola>\,  is  sometimes  called  the  Boscovich 
ratio,  but  is  generally  known  as  the  eccentricity  and  is  represented 
by  e. 


THE  HYPERBOLA. 


275 


172 

and  compare  it  with 


But 


Problems.  —  1.    To  find  an  expression  for  the  ratio 

PF  FM' 

pm' 

PF>  ^  P]i[  ^  r.V"  ^TA  +  .LV 
PM'      QD' 

TA   ^  QA_  Qj.  

^.V"      AD''  AX" 

TA  +  AX"  ^  AX"  _  AX_  PF 
~  AD' 
^PF 

pm' 


QA  +  AD' 
PF 
PM' 


QD'       QA  +  AD' 
TA  +  AX'  _  QA  +  AD' 

Ajy     ' 


AD      FM 


P'F 


2.    To  find  an  expression  for  the  ratio  and  compare  it 

.  ,     PP  F'M' 

with 


PF 

pm' 


But 


Hence 


FF'  _ 
F'M' 
TA' 
A'X' 
TA'  +  A'X' , 
Q'A'  +  A'D' 
FF 


F'B"  ^  rX'  ^  TA'  +  A'X' 

Q'D'       Q'D'      Q'A'  +  A'D'' 

Q'A'         TA'  +  A'X'      Q'A'  +  A'D' 
-i .  or =  -^ » 


A'D'" 
AJX 
A'D' 
PF 

FM 


A'X' 


A'D' 
PF 


^^(§65,  Ex.  2)  = 

AD^^  '      FM 


FM' 

3.  Draw  FF  and  show  that 

FF  ^  FF  ^FF  ^FF  ^ 
(  FM     FM'      FM'      FM 

4.  Show  by  Fig.   302   that  PF  -  FF  is   constant,    and    that 
FF  —  FF  equals  the  same  constant. 


■    ~-~--^p' 

M' 

M                P^^- 

^^^ 

^xr 

Zy^ 

pyA! 

D'                  D 

2\~F 

Fig.  365. 

F  and  F  are  called  foci,  and  DM  and  D'M*  are  called  directrices. 


276 


ELEMENTS   OF   GEOMETRY. 


6.   Find  a  line  that  shall  represent  the  constant,  PF'  —  PF. 
A  and  J!  being  points  of  the  curve, 

APi-AF  =K 

A'F-A'F  =  K 


AA'  +  AF 


-A'F'  =  Ki 


2  A'F  -2AF  =  0, 
A'F'  =  AF. 


2  AA'  =  2K 
or  K=AA'. 

In  addition  to  determining  the  fact  that  PJ''— PF=A4',  we 
have  determined  that  A'F'  =  AF. 

6.  Show  that  A'D'  =  AD,  and  that  the  hyperbola  is  symmet- 
rical with  respect  to  FF'  and  also  with  respect  to  the  perpendicular 
bisector  of  FF'. 

Note.  —  Because  of  the  relation  developed  in  I*rob.  4,  the 
hyperbola  may  be  defined  as:  The  locus  of  a  point  moving  in  a 
plane  so  that  the  difference  of  its  distances  from  two  fixed  points, 
also  in  the  plane,  will  remain  fixed. 

If  the  condition  that  the  locus  be  a  plane  curve  be  removed, 
the  locus  would  be  a  hyperboloid  of  revolution  of  two  nappes. 
If  revolved  on  the  vertical  axis  of  symmetry,  a  single  sheet  will 
be  generated,  which  would,  upon  investigation,  turn  out  to  be  a 
warped  surface  as  well  as  a  surface  of  revolution,  and  one  that 
could  be  generated  by 
a  straight  line  revolv- 
ing about  another 
straight  line,  not  in 
the  same  plane  with 
it. 


7.  Construct  a 
plane  curve  such  that 
the    ratio  of  the  dis- 


F'  F 

Fig.  366. 

tances  of  all  points  of  the  curve  from  a  fixed  point  and  from  a 
fixed  straight  line  shall  be  |. 

8.   Show  how  to  construct  a  hyperbola  by  a  continuous  motion. 


THE   HYPERBOLA. 


277 


173,  Theorem  I.  If  any  x>oint  (H)  be  on  the  con- 
cave side  of  a  branch  of  a  hyperbola,  i.e.  within  the  cone, 
HF'-  HF>  AA'. 


Fig.  367. 


PF'  -PF  =  AA 
PH^PH 

PF'  -\-PH-PF  =  AA'  -f  PH 
FH-PF<PH 

.-.  HF'-FH>AA' 


Q.  E.  D. 


Theorem  II.  If  any  point  (II)  be  on  the  convex  sides 
of  both  branches  of  a  hyperbola,  i.e.  not  within  the  cone, 
irF'-HF<AA'. 


PF'  -  PF 

pir 


AA' 
PIT 


PF'  -  PH'  -PF  =  AA'-  Pir 
H'F'  -  PF'  <  PH' 

H'F'  -  PH  -PF<AA' 
HF'-(Pir  +  PF)  <  A  A'. 
.'.  HF'-HF<AA'. 


Q.  E.  D. 


278 


ELEMENTS    OF   GEOMETRY. 


Theorkm  III.  If  a  straight  line  be  drawn  through  a 
point  of  a  hyperbola  so  as  to  make  equal  angles  with  tJie 
focal  radii  to  the  same  point,  it  will  be  a  tangent  to  the  curve. 

Analysis.  —  If  the  straight  line  TP  touch  the  hyperbola 
at  P,  and  every  other  point  of  TP  be  exterior  to  the 
curve,  it  will  be  a  tangent. 


Demonstration. 
the  Z  FPF. 


■The  line  TP,  by  hypothesis,  bisects 

J/ 


Fig.  368. 


If  FG  be  drawn  per- 
pendicular to  TP,  GPF 
will  be  isosceles,  TP 
will  be  the  perpendic- 
ular bisector  of  FG, 
and  since  PF' -  PF  = 
AA',  PF'-PG  =  AA' 
ovFG  =  AA'{2a). 

If  any   point   of  the 
angle  bisector,  as  J,  be  joined  with  F,  G,  and  F,  we  will 
have, 

JF-JF=,rF-JG<2a. 

Hence  J  (any  point  other  than  P)  is  exterior  to  the 
curve  and  TP  is  a  tangent  at  P.  q.  e.  d. 

Remark. — A  normal  would  bisect  the  adjacent  angle  formed  by 
the  focal  radii  to  the  point  of  tangency. 

Note.  —  In  the  parabola  where  there  is  but  one  focus,  the  other 
may  be  said  to  be  on  the  axis  at  infinity.  Tliis  view  makes  pos- 
sible the  general  enunciation.  Lines  drawn  from  the  foci  of  a 
conic  section  to  any  point  of  the  curve  make  equal  angles  with  the 
tangent  at  that  point. 


THE   HYPKRBOLA. 


279 


Fig.  369. 


174.    Since   the   tangent   at  P  bisects  the  ZF'PF,  it 
separates    F'F    into 

segments   proper-       \  p^ 

tional  to  the  adjacent 
sides.  .'.  F'Tis  the 
greater  segment  and 
the  foot  of  a  tangent 
"will  always  fall  with- 
in^'A 

As  the  point  of 
tangency  (P)  becomes 
more  remote  from  A,  the  point  T  approaches  G,  bxit  does 
not  reach  it  until  P  shall  have  passed  to  infinity;  in 
which  case  F'P,  TP,  and  FP  will  be  parallel.  In  this 
limiting  position  the  tangent  is  called  an  asymptote.  It  is 
a  determined  line  toward  which  the  curve  approaches, 
and  to  which  it  is  tangent  at  an  infinite  (co)  distance 
from  C. 

In  §  173  it  is  shown  that  a  perpendicular  from  F  to  a, 
tangent  A\-ill,  if  pro- 
duced its  length,  have 
its  extremity  in  the 
line  dra"\vn  from  F'  to 
the  point  of  tangency. 

Because  F'P  and 
CP  are  paral]el,  the 
ZF'GF  will  be  90° 
and  its  vertex  must  ^^^-  ^'^■ 

be  in  the  circumference  of  a  circle  on  FF'  as  a  diameter. 


It  was  also  shown  that  F'G  =  2a;  and  G  must  be  in. 


280 


ELEMENTS   OF   GEOMETRY. 


the  circumference,  having  P  for  its  centre  and  2  a  for  its 
radius. 

Hence,  to  construct  an  asymptote  : 

On  FF'  as  a  diameter,  construct  a  circumference. 

With  F  as  a  centre  and  A  A'  as  a  radius,  construct  an 
arc  intersecting  the  preceding  one  at  G  and  G\ 

Through  C  draw  lines  parallel  to  FG  and  to  FG\ 
They  will  be  the  asymptotes. 

If  the  construction  be  made  from  F  instead  of  from 
F,  it  would  be  found  that  the  asymptotes  to  one  branch 
are  also  asymptotes  to  the  other. 

If  at  ^  a  perpendicular  be  erected,  AK  —  FS, 
CK  =  CF.  If  CA  be  represented  by  a,  CF  by  c,  and 
AK  by  6,  we  have  (^z=a^  -{-  Ir. 

The  asymptotes  are  the  diagonals  of  a  rectangle,  the 
sides  of  which  are  2  a  and  2  h. 


Fio.  8T1. 


Exercises.  —  1.  Show  how  to  draw  a  tangent  to  the  hyperbola 
at  a  given  point  on  the  curve. 

2.  Show  how  to  draw  a  tangent  to  a  hyperbola  from  any  point. 
What  limitations  are  there  as  to  the  position  of  the  point  from 
which  the  tangent  is  to  be  drawn  ?  Show  that  in  general  there 
may  be  two  tangents. 


THE   HYPEKBOLA.  281 


PARTICULAR  CASES. 

175.  If  the  plane  which  cuts  a  hyperbola  from  a  cone 
be  moved  parallel  to  itself  and  toward  the  vertex,  the 
vertices  of  the  curve  and  the  foci  approach  each  other, 
and  Avhen  the  plane  passes  through  the  vertex  of  the 
cone,  the  vertices  and  foci  will  all  coincide  and  the 
hyperbola  will  become  two  intersecting  straight  lines. 

If  the  plane  be  moved  parallel  to  itself  and  atvay  from 
the  vertex  of  the  cone,  the  vertices  and  the  foci  of  the 
curve  will  recede  from  each  other,  the  curvature  in  the 
neighbourhood  of  the  vertices  will  decrease,  and  the  hyper- 
bola will  approach  two  parallel  lines,  both  at  infinity  and 
at  an  infinite  distance  from  each  other. 

If  the  cone  with  a  given  circular  base  should  approach 
a  cylinder,  the  section  which  yields  a  hyperbola  will 
approach  parallel  straight  lines. 

If  the  angle  which  the  elements  make  with  the  axis 
should  approach  90°,  the  vertices  of  the  two  branches 
would  approach  each  other,  the  curvature  would  dimin- 
ish, and  when  the  angle  should  become  90°,  the  cone 
would  form  a  plane  and  the  two  branches  of  the  hyper- 
bola would  fall  together  in  a  single  straight  line. 

Note.  —  Recalling  the  figures  wherein  the  sections  of  a  cone 
are  represented,  and  conceiving  a  perpendicular  to  the  plane  of  the 
picture  as  erected  at  A,  we  may  by  passing  a  plane  through  this 
perpendicular,  and  rotating  it  about  the  perpendicular  as  an  axis, 
obtain  the  different  sections  in  turn. 

Beginning  in  such  a  position  that  the  section  shall  be  a  hyper- 
bola, and  rotating  toward  the  position  which  will  give  a  parabola, 
we  see  that  the  vertex  and  the  focus  of  the  second  branch  recede 
from  the  first,  and  that  when  the  plane  becomes  parallel  to  a  tan- 


282  ELEMENTS   OF   GEOMETRY. 

gent  plane,  the  second  vertex  and  its  adjacent  focus  will  have 
passed  to  infinity. 

The  instant  that  the  rotation  carries  the  secant  plane  beyond 
the  position  when  it  gives  a  parabola,  we  get  an  ellipse  with  the 
second  focus  and  vertex  at  a  very  great  distance  from  the  fii^st ; 
as  the  rotation  continues,  the  foci  approach  each  other,  and  when 
the  rotating  plane  is  perpendicular  to  the  axis  the  foci  will  have 
coincided  and  the  section  will  be  a  circle,  i.e.  a  particular  case  of 
an  ellipse. 

If  the  rotation  be  continued,  the  foci  will  again  separate  and 
the  ellipse  become  narrower  and  narrower  until  the  secant  plane 
becomes  a  tangent  plane,  when  the  ellipse  will  have  degenerated 
to  a  right  line,  which  is  also  a  particular  case  of  the  parabola. 

This  last  position  might  have  been  reached  by  rotating  the 
secant  plane  the  reverse  way,  in  which  case  the  straigTit  line  would 
be  the  limit  toward  which  the  hyperbola  would  approach. 

The  properties  of  the  ellipse  and  hyperbola  are  complementary 
and  may  be  deduced  together. 

One  conic  section  may  be  projected  into  another  by  considering 
some  point  not  in  the  plane  of  the  section  as  the  source  of  light, 
and  the  shadow  be  intersected  by  planes  occupying  different 
positions. 


SOME   PROBLEMS   IN  SOLID   AND   SPHER- 
ICAL  GEOMETRY. 

1.  Show  that  if  a  straight  line  be  perpendicular  to  a  plane  its 
projection  on  any  other  plane  will  be 
perpendicular  to  the  line   of   intersec- 
tion of  the  two  planes. 

2.  Show  how  to  draw  a  straight  line 
through  the  vertex  of  a  triedral  so  as  to 
separate  it  into  three  isosceles  triedrals. 

3.  Show  how  to  circumscribe  a  circle 
about  a  spherical  triangle. 

4.  Show  how  to  inscribe  a  circle  ^^^  3^2 
within  a  spherical  triangle. 

5.  Through  a  point  on  the  surface  of  a  sphere  to  draw  an  arc 
of  a  great  circle  that  shall  be  tangent  to  a  given  small  circle. 

6.  Use  concentric  spheres  to  show  that  the  intensity  of  a  light 
varies  inversely  as  the  square  of  the  distance  from  the  source. 

7.  The  three  angles  of  a  spherical  triangle  are  64°,  85°,  and 
122°,  on  a  sphere  the  diameter  of  which  is  18  inches ;  find  the 
area. 

8.  Eegarding  the  eartli  as  a  sphere  the  radius  of  which  is 
3956  miles,  find  the  area  of  a  trirectangular  spherical  triangle. 

9.  Assume  numerical  values  for  each  of  the  parts  which  are 
sufiicient  to  determine  a  spherical  triangle  and  make  the  con- 
struction on  a  spherical  blackboard. 

10.  A  sphere,  the  radius  of  which  is  1,  is  circumscribed  by  a 
cube,  a  cylindei',  and  a  cone,  the  elements  of  which  make  an  angle 
of  30°  with  the  axis ;  find  the  relative  surfaces  and  volumes. 

283 


284 


ELEMENTS   OF   GEOMETKY. 


11.  A  sphere  is  circumscribed  by  a  cylinder  ;  the  lower  base  of 
the  cylinder  is  the  base  of  a  right  circular  cone,  the  vertex  of  which 
is  at  the  centre  of  the  upper  base  ;  find  the  volume  which  is  com- 
mon to  the  cone  and  the  sphere. 

12.  To  find  approximately  the  distance  of  the  horizon  on  the 
ocean  as  seen  from  an  elevation. 

e  (c  +  2  iZ)  =  (P. 

For  any  elevations  that  exist  on  the  earth,  the  e  that  is  added 
to  2  jB  may  be  neglected  without  serious  error. 

The  e  which  is  a  factor  cannot  be  neglected. 

.-.  c  =  —  or  (P  =  2  eB. 
2B 

If  B  be  4000  miles, 

e  =  j(nnj*'^  (in  miles). 

66 


In  feet, 


e  =  -=-  X  6280  =  ^  (P 
8000  800 


d^=-  (P. 
100  3 


.".  e  (in  feet)  =  |(P  in  miles. 
If  d  be  one  mile,  c= |  of  a  foot,  or  8  inches. 


¥ni.  3V8. 


13.  Mt,  Washington  is  6288  feet  high ;  how  far  is  the  ocean 
horizon  from  it  ? 

14.  Standing  on  the  deck  of  a  ship  is  a  man  whose  eyes  are  20 
feet  from  the  water.  A  second  ship,  the  hull  of  which  stands  12 
feet  out  of  water,  is  seen  to  be  just  "hull  down."  What  is  the 
distance  apart  of  the  two  ships  ? 

16.  Two  triedrals  have  two  facial  angles  of  one  equal  to  two 
facial  angles  of  the  other,  but  the  enclosed  diedrals  are  not  equal ; 
what  relations  exist  between  the  third  facial  angles  in  each  ? 

16.  Show  that  the  three  planes  which  bisect  the  diedrals  of  a 
triedral,  pass  through  one  line. 

17.  Show  that  the  three  planes  passed  through  the  bisectors  of 
the  facial  angles,  perpendicular  to  those  faces,  pass  through  one  line. 


PBOBLEMS.  285 


PROBLEMS  IN  LOCI. 

18.  Find  the  locus  of  points  in  space  which  are  equally  distant 
from  two  given  points. 

19.  Find  the  locus  of  points  which  are  equally  distant  from 
three  given  points. 

20.  Find  the  locus  of  points  which  are  equally  distant  from  two 
given  planes. 

21.  Find  the  locus  of  points  which  are  equally  distant  from 
three  given  planes. 

22.  Find  the  locus  of  a  point  which  moves  so  that  its  distances 
from  the  three  edges  of  a  triedral  are  always  equal  to  each  other. 

23.  Find  the  locus  of  points  in  a  given  plane  which  are  equally 
distant  from  two  points  which  may  or  may  not  be  in  the  given 
plane. 

24.  Find  the  locus  of  a  point  which  moves  so  that  the  ratio  of 
its  distances  from  two  parallel  lines  always  equals  1. 

25.  Find  the  locus  of  a  point  which  moves  so  that  the  ratio  of 
its  distances  from  a  fixed  line  and  a  fixed  point  of  that  line  is 
constant. 

26.  Find  the  locus  of  points  which  are  the  centres  of  the  sections 
of  a  sphere  made  by  planes,  (a)  Which  pass  through  a  fixed  line. 
(6)  Which  pass  through  a  fixed  point. 

27.  Find  the  locus  of  the  centre  of  a  sphere  which  is  tangent  to 
three  given  planes. 

SOME  PROBLEMS  INVOLVING  THE  INTERSECTION  OF 
LOCL 

28.  Find  the  locus  of  a  point  which  moves  so  that  the  ratio  of 
its  distances  from  two  points  equals  1  ;  and  the  ratio  of  its  dis- 
tances from  two  other  points  equals  1. 

29.  Find  the  locus  of  points  which  are  at  a  given  distance  from 
P,  and  twice  as  far  from  Q. 


286  ELEMENTS   OF   GEOMETRY. 

30.  Find  the  locus  of  points  which  are  at  a  given  distance,  and 
the  ratio  of  the  distances  of  which  from  K  and  Q  equals  1. 

31.  Locate  a  point  which  shall  be  equally  distant  from  two  given 
planes  and  from  a  given  point. 

32.  Locate  a  point,  the  ratio  of  the  distances  of  which  from  two 
planes  equals  1,  and  the  ratio  of  its  distances  from  two  points 
equals  1. 

33.  Find  the  locus  of  a  point  which  moves  so  that  it  remains  at 
a  fixed  distance  from  a  given  straight  line,  and  the  ratio  of  its  dis- 
tances from  two  fixed  points  equals  1. 

34.  Locate  a  point  so  that  the  difference  of  the  squares  of  its 
distances  from  two  given  points  is  constant. 

36.  Find  the  locus  of  points  which  are  equally  distant  from  the 
surface  of  a  sphere,  and  which  are  so  situated  that  the  ratio  of 
their  distances  from  two  given  points  is  L 

36.  Locate  a  point  in  a  plane  so  that  the  sum  of  the  squares  of 
its  distances  from  two  given  points  not  in  the  plane,  is  fixed ;  and 
its  distance  from  a  given  line  of  the  plane  is  also  fixed. 

37.  Find  the  locus  of  a  point  from  which  two  spheres  subtend 
equal  visual  cones,  and  a  given  segment  of  a  straight  line  subtends 
a  given  angle. 

THE  FIVE   REGULAR   POLYEDRONS. 

38.  Show  that  four  equal  equilateral  triangles  may  be  placed  so 
as  to  enclose  a  volume. 

Definition. — The  figure  thus  formed  is  called  a  regular  tetraedron. 

39.  One  edge  of  a  tetraedron  is  a  ;  deduce  formulae  V    ^/Tr^-y 
for  its  superficial  area  and  for  its  volume.  \__k__A 

40.  Show  that  six  equal  squares  may  be  placed  so  Fio-  874. 
as  to  enclose  a  volume. 

Definition.  —  The  figure  thus  formed  is  called  a  regular  Hexa- 
edron  or  Cube. 

Remark. — The  superficial  area  and  the  volume  we  are  already 
familiar  with. 


PROBLEMS.  287 

Note.  — It  is  recommended  to  the  student  that  he  make  paste- 
board models  of  the  regular  polyedra. 

41.  Show  that  eight  equal  equilateral  triangles  may  be  so  placed 
as  to  enclose  a  volume. 

Definition.  —  The  figure  thus  formed  is  called  a 
regular  octaedron. 

Fig.  8V5. 

42.  One  edge  of  a  regular  octaedron  is  a  ;  deduce 

formulae  for  the  superficial  area  and  the  volume. 

43.  Show  that  twelve  regular  pentagons  may  be  so  placed  as  to 
enclose  a  volume. 

Definition.  —  The  figure  thus  formed  is  called  a  regular  do- 
decaedron. 


Fig.  376. 

44.  One  edge  of  a  dodecaedron  is  a ;  deduce  formulae  for  the 
superficial  area  and  the  volume. 

Bemark.  —  The  student  who  solves  this  problem  without  assis- 
tance is  presumed  to  have  a  pretty  good  working  power  in 
geometry. 

45.  Show  that  ticenty  equal  equilateral  triangles  may  be  so 
placed  as  to  enclose  a  volume. 

Definition. — The  figure  thus  formed  is  called  a  regular  icosaedron. 

46.  One  edge  of  a  regular  icosaedron  is  a  ;  deduce  formulae  for 
the  superficial  area  and  the  volume. 

47.  Show  that  the  five  regular  polyedra  already  named  are  the 
only  regular  polyedra  that  can  be  formed. 

Note.  —  There  are  many  combinations  of  polygons  that  may 
be  used  to  enclose  a  volume  ;  as  one  may  see  by  looking  over  the 
sketches  of  any  work  on  crystallography. 


288  ELEMENTS   OF   GEOMETRY. 


NUMERICAL  EXERCISES. 

48.  An  octagonal  building  is  80  feet  across  (from  face  to  face); 
the  roof  is  to  be  a  hip  roof,  and  one-third  pitch  ;  *  find  the  length 
of  the  hip  rafters. 

49.  A  cylindrical  cistern  is  to  have  an  inside  diameter  of  20  feet 
and  a  depth  of  10  feet,  and  is  to  be  made  of  concrete  1  foot  thick  on 
the  bottom  ;  the  wall  is  to  be  2  feet  thick  at  the  bottom  and  1  foot 
thick  at  the  top,  the  slope  being  uniform.  Each  cubic  foot  costs 
30  cents  in  place.    What  will  be  the  cost  of  the  cistern  ? 

60.  The  chimney,  160  feet  high,  at  a  power  house,  is  a  friistum 
of  a  square  pyramid  16  feet  square  at  the  base,  and  6  feet  square 
at  the  top  ;  the  flue  is  3  feet  square  throughout  the  entire  length. 
Estimating  19  bricks  to  the  cubic  foot,  how  many  will  be  required 
to  put  up  the  chimney  ? 

61.  Before  any  spoliation,  the  great  pyramid  in  Egypt  had  for 
its  base  a  square  each  side  of  which  was  763.8  feet  and  its  altitude 
was  486.26  feet ;  what  was  its  volume  in  cubic  yards  ? 

62.  A  cask  3  feet  high  has  a  bung  diameter  of  3.18  feet,  and  a 
head  diameter  of  2.55  feet.  A  gallon  is  231  cubic  inches  ;  find  the 
number  of  gallons. 

63.  It  is  proposed  to  put  a  dam  across  a  canon  for  the  purpose 
of  making  a  reservoir. 

The  slope  of  the  hill  on  one  side 
is  1  on  3  ;  on  the  other  1  on  1.    The      iF^^^^^^I^:-^^       y^l 
dirt  on  the  side  of  the  dam  away  from 
the  water  will  lie  at  a  slope  of  1  on  ^'o.  .S78. 

1,  while  on  the  side  toward  the  water  the  slope  will  be  1  on  4. 

The  dam  is  to  be  90  feet  high  with  a  roadway  12  feet  wide 
on  top. 


^ 


*  Note.  —  One-third  pitch  indicates  a  slope 

of  roof  such  that  a  —  -.    The  /.  0  wQl  be  some 
3 

what  greater  than  33°  41'.  P^^  ^^^ 


h 


PROBLEMS.  289 

It  will  cost  30  cents  per  cubic  yard  to  place  the  dirt  in  the  dam. 
What  will  the  dam  cost  if  built  ? 

54.  A  certain  reservoir  is  an  inverted  frustum  of  a  cone ;  the 
upper  circumference  is  355  metres ;  the  lower  circumference  is 
120  metres  ;  and  the  length  of  the  slope  is  30  metres.  Find  the 
capacity  of  the  reservoir  in  hectolitres. 

55.  The  brick  to  be  used  in  lining  a  well  are  8"  (inches)  x  4" 
X  2|".  The  well  is  to  be  4'  (feet)  in  the  clear,  and  60'  deep. 
How  many  brick  will  be  required  ? 

56.  A  barrel  contains  31|^  gallons  ;  its  bung  diameter  is  18  inches, 
and  its  head  diameter  is  15  inches.     What  must  be  its  length  ? 


THEOREMS. 

57.  If  two  tetraedrons  have  a  triedral  in  each  equal,  the  ratio 
of  their  volumes  equals  the  ratio  of  the  products  of  the  three  edges 
which  determine  the  angle. 

58.  The  planes  which  pass  through  the  concurrent  edges  of  a 
tetraedron  and  the  middle  points  of  the  opposite  edges,  all  pass 
through  one  line. 

59.  The  lines  joining  the  vertices  of  a  tetraedon  with  the  points 
of  intersection  of  the  medians  of  the  opposite  faces,  are  concurrent ; 
and  each  is  separated    into  parts 

which  have  the  ratio  of  3  : 1. 

Note.  —  This  point  is  the  centre 
of  gravity  of  the  tetraedron. 

60.  The  lines  joining  the  middle 

points  of  the  opposite  edges  of  a 

tetraedron  are  concurrent ;  and  the 

point  is  the  centre  of  gravity. 

^  o  J  Fio.  879. 

61.  The    altitude    of    a  regular 

tetraedon  equals  the  sum  of  the  perpendiculars  on  the  four  faces, 
from  any  point. 


290 


ELEMENTS    OF   GEOMETRY. 


FiQ.  880. 


62.  If  a,  b,  c,  and  d  represent  the  altitudes  of  a  tetraedron, 
and  a',  b',  c',  and  d'  the  parallel  perpendiculars  to  the  faces  from 
any  point,  then : 

abed 

68.  A  plane  angle  may  be  projected  into  an  equal  angle,  a 
smaller  angle,  or  a  greater  angle. 

64.  Any  plane  area  orthogonally  projected  on  a  plane,  will  bear 
to  the  projected  area  the  same 
ratio  that  a  straight  line  in  the 
given  plane  area  perpendicular 
to  the  intersection  of  the  two 
planes,  bears  to  the  projection 
of  that  line. 

Note.  —  The  ratio  of  the  pro- 
jected segment  of  a  line  divided 
by  the  segment  projected  is 
called  the  cosine  of  the  angle 
of  inclination  of  the  two  lines,  as  the  student  will  see  when  he 
begins  the  study  of  Trigonometry. 

It  is  customary  to  say  "  a  projected  plane  area  equals  the  given 
area  multiplied  by  the  cosine  of  the  angle  of  inclination." 

65.  If  a  tetraedron  is  cut  by  a  plane  parallel  to  a  pair  of  oppo- 
site edges,  the  section  will  be  a  parallelogram,  the  area  of  which 
will  be  a  maximum  when  the  parallelogram  is  a  middle  section. 

66.  The  six  planes  which  bisect  the  six  diedrals  of  a  tetraedron 
all  pass  through  one  point. 

67.  Through  any  four  points,  one,  and  only  one,  sphere  may  be 
passed. 

68.  If  a  circle  be  circumscribed  about  each  face  of  a  tetraedron 
and  perpendiculars  be  erected  at  the  centres,  they  will  be  concurrent. 

69.  Three  spheres  intersect  each  other ;  the  planes  of  their  in- 
tersection pass  through  a  common  line ;  and  the  tangents  to  the 
tliree  spheres  from  any  point  in  this  line  will  be  equal  to  each 
other. 


PROBLEMS.  291 


PROBLEMS. 

70.  The  same  number  expresses  the  volume  and  the  surface  of 
a  sphere.     What  is  its  radius  ? 

71.  Show  that  the  surface  generated  by  a  triangle  revolving 
about  one  side  equals  the  length  of  the  two  generating  sides,  mul- 
tiplied by  the  path  described  by  the  centre  of  gravity. 

72.  Show  that  the  volume  generated  by  a  triangle  revolving 
about  one  side,  equals  the  area  of  the  triangle,  multiplied  by  the 
distance  traversed  by  the  centre  of  gravity. 

73.  Tillamook  light  is  138  feet  above  high  water ;  the  lookout 
on  board  ship  is  60  feet  from  the  water  ;  the  tide  is  12  feet.  What 
is  the  difference  of  the  distances  at  which  the  light  can  be  seen  from 
the  ship  at  high  tide  and  at  low  tide  ? 

74.  Find  the  polyedron  that  would  be  formed  by  joining  the 
middle  points  of  the  faces  of : 

(a)  A  tetraedron,  (c)  An  octaedron, 

(b)  A  cube,  ((Z)  A  dodecaedron, 

(e)  An  icosaedron. 

75.  Show  how  to  find  approximately  the  diameter  of  the  earth 
by  comparing  the  angles  of  elevation  of  ^p 
the  north  pole  as  observed  at  two  points, 
A  and  B,  which  are  on  the  same  meridian 
and  a  known  distance  from  each  other. 

Note. — The  distances  of  the  fixed  stars 
from  the  earth  are  so  great  that  lines  from 
any  one  of  them  to  different  points  on  the 
earth  are  practically  parallel. 

76.  Show  that  if  through  any  point  on 
the  surface  of  a  sphere,  three  chords  be  ■^'®-  ^®^' 
drawn  at  right  angles  to  each  other,  the  sum  of  the  squares  of  these 
chords  equals  the  square  of  the  diameter. 

77.  A  sphere  8  inches  in  diameter  has  a  cylindrical  hole  4 
inches  in  diameter  bored  through  it,  the  axis  of  the  cylinder  passing 
through  the  centre  of  the  sphere.    What  is  the  volume  that  remains? 


292  ELEMENTS   OF   GEOMETRY. 

78.  Having  given  the  radius  of  the  inscribed  sphere,  find  the 
volumes  of : 

(a)  The  circumscribed  regular  tetraedron, 
(6)  The  circumscribed  regular  hexaedron, 

(c)  The  circumscribed  regular  octaedron, 

(d)  The  circumscribed  regular  dodecaedron, 

(e)  The  circumscribed  regular  icosaedron. 

79.  Show  geometrically  that : 

(a  +  6)8  =  a^  +  3a26  +  Sab^-  +  6». 

80.  Show  that  the  area  of  a  zone  of  one  base  equals  the  area  of 
the  circle  which  has  the  chord  of  the  generating  arc  for  its  radius. 

81.  Show  that  the  volume  between  the  surfaces  of  two  concen- 
tric spheres  (a  spherical  shell)  equals  the  frustum  of  a  right  cir- 
cular cone,  the  lower  base  of  which  has  for  its  radius,  the  radius  of 
the  outer  sphere  ;  for  the  radius  of  the  upper  base,  the  radius  of  the 
inner  sphere  ;  and  for  its  altitude,  four  times  the  difference  of  the 
radii. 

82.  Show  that  when  the  frustum  of  a  pyramid  or  cone  is  being 
considered : 

6  3 

88.  Apply  the  prismoidal  formula  to  the  determination  of  the 
volumes  of  prolate  and  oblate  spheroids. 

84.  An  elliptical  reservoir  has  for  its  major  and  minor  diameters 
200  and  100  feet  respectively  ;  the 
uniform  depth  is  20  feet.     What 
will  be  the  contents  in  gallons  ? 

85.  A  vault  has  a  rectangular 
floor  100  X  30  feet,  and  an  arched 
ceiling  15  feet  high,  which  comes 

Fio    882 

down  to  the  floor ;  the  arch  is  para- 
bolic ;  the  ends  are  plane.     Find  the  cubic  contents. 


PROBLEMS.  293 


LOCI. 

86.  Find  the  locus  of  the  vertex  of  spherical  triangle,  the  verti- 
cal angle  being  equal  to  the  sum  of  tlie  other  two. 

87.  Find  the  locus  of  points  in  space,  such  that  the  ratio  of 
their  distances  from  two  fixed  points  is  constant. 

Note. — This  is  the  problem  of  the  lights  not  confined  to  a  plane. 

88.  Find  the  locus  of  a  point  in  space  which  moves  so  that  the 
sum  of  its  distances  from  two  fixed  points  always  remains  the  same. 

89.  Find  the  locus  of  the  centre  of  a  sphere  which  is  tangent  to 
a  given  plane,  and  the  surface  of  which  passes  through  a  fixed 
point. 

90.  Find  the  locus  of  a  point  which  moves  so  that  the  ratio  of 
its  distances  from  a  fixed  point  and  from  a  fixed  plane  is  constant 
and  less  than  1. 

91.  Find  the  locus  of  points,  the  ratio  of  the  distances  of  which 
from  a  given  cylinder  and  from  a  given  circumference,  is  1 ;  the 
centre  of  the  circle  being  in  the  axis  of  the  cylinder,  the  radius 
greater  than  that  of  the  cylinder,  and  the  plane  perpendicular  to 
the  axis  of  the  cylinder. 


THE   END. 


00  2  1      8 


AT 

XX)S  angklbs 

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UNIVERSITY  OF  CALIFORNIA,  LOS  ANGELES 

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